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Full text of "Genetics"

RWIN J. HERSKOWITZ 







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Second Edition 




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GENETICS 



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IRWIN H. HERSKOWITZ 

Hunter College, 
The City University of New York 



Second Edition 



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LITTLE, BROWN AND COMPANY 



Boston and Toronto 



COPYRIGH1 j 1962, 1965, Bl LITTLE, BROWN AND COMPANY Mm.) 

\1 I RK.IIIs Kl si RVFD. NO I'ART OF THIS BOOK 
\1 \1 HI REPRODUCED IN ANY FORM WITHOUT 
I'l RM1SSION IN WRITING FROM THI PUBLISHER. 

LIBRARY OF CONGRESS CATALOG CARD NO. 65-17335 



FIRS I I'RIN LING 



Published simultaneously in Canada 
by Little, Brown & Company {Canada) Limited 

PRINTED IN THE UNITED STATES OF AMERICA 



PREFACE 



S 



Iince the beginning of this cen- 
tury the science of genetics 
has had a spectacular growth. 
The discovery of basic principles and the 
application of these principles are occurring 
at an ever-increasing rate. It is generally 
agreed that a knowledge of genetics is essen- 
tial for an understanding of present and fu- 
ture biology. The impact of genetics is not 
restricted, however, to professional teachers 
and research workers in pure and applied 
biology, nor to physicians and dentists. 
More and more students whose major in- 
terest is in psychology, biochemistry, chem- 
istry, biophysics, physics, or mathematics 
find that the study of genetics offers new and 
challenging opportunities in these various 
fields. 

How can a text for an introductory course 
in genetics be best organized to serve stu- 
dents with such varied interests? An intro- 
ductory text must provide the reader with an 
understanding of the nature of the genetic 
material, for this knowledge is a prerequisite 
for a fruitful genetic approach to the solution 
of problems in biology and all the other 
fields mentioned. Accordingly, insofar as 
possible, the subject matter of this book is 
arranged so that principles dealing with the 
nature of the genetic material are separated 
from the applications of these principles. 
The nature of the genetic material is studied 
through the use of the operations or methods 
of recombination, mutation, chemistry and 
physics, replication, and function. The pres- 
entation is designed to encourage the reader 
to use his powers of inductive reasoning to 
arrive at the primary generalizations of 



genetics on the basis of experimental evi- 
dence. Whenever feasible, genetic principles 
are derived scientifically — by recognizing 
and stating a problem, designing appropriate 
experiments to test hypotheses, analyzing the 
experimental results, and drawing conclu- 
sions. The aim is to present genetics as a 
rational, organized body of knowledge. 

Because of its importance the introductory 
genetics course is being offered more fre- 
quently in the earlier rather than the later 
years of college study. Since such a course 
is elected more and more frequently by 
students who do not wish to specialize in 
biology, simple biological examples and ter- 
minology are used whenever possible, and 
certain biological phenomena generally un- 
derstood by students specializing in biology 
are explained in some detail. Because many 
students in a first course in genetics may not 
have an adequate background, certain as- 
pects of chemistry and physics important 
for understanding genetics are described in 
greater detail than in other texts. 

No single text can include the ways each 
principle of genetics apply to every plant 
and animal studied, or give examples of the 
application of each of these principles to all 
the different kinds of organisms. Accord- 
ingly, only one or a few experimentally 
favorable or historically important organisms 
are usually employed in this text to establish 
a principle or to illustrate an application. 
Additional proofs, applications, or examples 
are left to the instructor who, depending 
upon his students' training and interest, can 
supply other illustrations by means of lec- 
tures and laboratory sessions or by means 
of assignments to detailed accounts in other 
texts and in the original literature. 

It is hoped that the text will stimulate 
readers to utilize the books and journals in 
their libraries. The reading of genetic works 
in the original after studying appropriate 
sections of the text can be a very rewarding 
experience. Accordingly, references requir- 



VI 



PREFACE 



Ing different degrees of sophistication are 
given at the ends of chapters. Since the later 

chapters deal with recent advances in genet- 
ics, whose discussion may be absent from 
already published textbooks, more references 
are given to particular workers in the later 
than in the earlier chapters. 

Part of a letter by G. Mendel and the 
Nobel Prize Lectures presented by geneticists 
are included in the book as supplements. 
These supplements should be completely un- 
derstandable, or nearly so. if appropriate 
chapters have been read beforehand, and can 
serve as a review and overview of genetic 
principles and their applications. The cita- 
tions to the literature included in the Nobel 
Prize talks should prove especially valuable 
to those who wish to do additional reading 
on key topics. The supplements can also 
function to bridge the gap between the text- 
book and the research worker, giving the 
reader some idea of the history of the subject 
and the personalities of the people involved. 

The Second Edition 

The subject matter is presented in forty-two 
chapters, each ending, as before, with a sum- 
mary, questions for discussion, and refer- 
ences. An appendix. "Elementary Biomet- 
rical Inferences," has been added, and the 
supplements now include three additional 
Nobel Prize lectures. A description of rep- 
resentative life cycles and genetic maps of 
several higher organisms have also been 
added. Recent advances — for example, in 
human, corn, and Drosophila genetics and in 



our understanding of the genetic code and 
the regulation of gene synthesis and gene 
action — have required that several chapters 
be combined or rearranged in sequence and 
that new chapters be written. Additional fig- 
ures, photographs, problems, and references 
are also included. 

Suggestions for Use of the Book 

The text contains more information than is 
usually covered in a one-semester, introduc- 
tory course for undergraduates. The chap- 
ters or chapter sections that are marked by 
an asterisk do not contain principles or 
terminology needed to understand unmarked 
parts and are, therefore, optional. 

A one-semester lecture course (meeting 
about 30 to 45 periods) can be based upon 
( 1 ) thirty chapters — those whose chapter 
numbers are unstarred, or (2) twenty-eight 
chapters — numbers 1 through 4 and 19 
through 42. 

A two-semester lecture course (meeting 
a total of about 60 to 90 periods) can be 
based upon (1) the first eighteen chapters 
for the first semester and the last twenty-four 
chapters for the second semester, or (2) all 
thirty unstarred chapters for the first semes- 
ter and all starred chapters and sections for 
the second semester. 

Acknowledgments 

I wish to thank my wife, Reida Postrel 
Herskowitz, for preparing the typescript, and 
my sons, Ira and Joel, and my present and 
former students for numerous suggestions. 



CONTENTS 

1 Genetic Material and Mitosis 1 

2 Meiosis and Chromosomal Segregation 75 

3 Segregation of Alleles 31 

4 Independent Recombination by Nonalleles 42 

5 Multiple Alleles; Multigenic Traits 57 
*6 Phenotypic Effects of Gene Action 69 

7 Sex Chromosomes and Sex-Linked Genes 90 

*8 Sex Determination 102 

9 Linkage and Crossing Over Between Genes 116 

10 Gene Arrangement; Crossover Maps 131 

11 Changes Involving Unbroken Chromosomes 149 

12 Structural Changes in Chromosomes 164 

*13 Radiation-Induced Structural Chromosome Changes 179 

*14 Point Mutations 189 

15 The Gene Pool; Equilibrium Factors 201 

*16 Genetic Loads and Their Population Effects 216 

*17 Chromosomal Rearrangements in Nature 228 

*18 Races and the Origin of Species 241 

19 Chemical Nature of Genes 252 

20 Organization and Replication of DNA in Vivo 265 

21 Replication of DNA in Vitro 279 

22 Clones; Transformation; Strand Recombination in Vitro 292 

23 Bacterial Mutation and Conjugation 306 

24 The Episome F 317 

25 Transduction 330 

26 Bacteriophage; Recombination and Genetic Maps 339 

27 Bacterial Episomes and Genetic Recombination 355 

28 RNA as Genetic Material 363 



%iii CONTENTS 

29 Extranuclear Genes 369 

30 The Genetic Control of Mutation 383 

31 The Molecular Basis of Mutation 391 

32 Gene Action and Polypeptides 404 

33 Polypeptide Synthesis and RNA 42J 

34 Genetic Amino Aeid Coding 436 

35 Regulation of Gene Synthesis 449 

36 Regulation of Gene Action — Operons 457 

37 Regulation of Gene Action — Gene Control Systems in Maize 465 

38 Regulation of Gene Action — Position Effect in Drosophila 473 

39 Regulation of Gene Action — Dosage Compensation 484 

40 Regulation of Gene Action — Its Molecular Basis in Higher 

Organisms 492 

41 Regulation of Gene Action — Growth, Differentiation, and 

Development 501 

42 The Origin and Evolution of Genetic Material 509 



Appendix — Elementary Biometrical Inferences 
Supplements 5-7 



519 



I 


Part of a Letter (1867 


II 


Nobel Prize Lecture ( 1 


III 


Nobel Prize Lecture ( 1 


IV 


Nobel Prize Lecture ( 1 


V 


Nobel Prize Lecture ( 1 


VI 


Nobel Prize Lecture ( 1 


VII 


Nobel Prize Lecture ( 1 


VIII 


Nobel Prize Lecture ( 1 


IX 


Nobel Prize Lecture ( 1 


X 


Nobel Prize Lecture ( 1 


Author Index 541 


Subject Index 546 



) from Gregor Mendel to C. Nageli s-9 

934) of Thomas Hunt Morgan s-15 
946) of Hermann Joseph Muller 5-79 
962) of Maurice H. F. Wilkins s-31 
959) of Arthur Romberg s-60 
958) of George Wells Beadle s-75 
958) of Edward Lawrie Tatum s-88 

958) of Joshua Lederberg s-98 
962) of James Dewey Watson 5-77/ 
962) of Francis H. C. Crick s-135 



The essential feature of the operational viewpoint is that an object 
or phenomenon under experimental investigation cannot usefully be 
defined in terms of assumed properties beyond experimental determi- 
nation but rather must be defined in terms of the actual operations that 
may be applied in dealing with it. . . . 

What is a gene in operational terms? 

L. J. Stadler, "The Gene," 
Science, 120:81 1-819, 1954 

Must we geneticists become bacteriologists, physiological chemists 
and physicists, simultaneously with being zoologists and botanists? 
Let us hope so. 

H. J. Muller, "Variation Due to Change in the Individual Gene," 

American Naturalist, 56:32-50, 1922 



GENETICS 



Chapter 1 

GENETIC MATERIAL 
AND MITOSIS 



S 



urely each of us has observed 
that we are the same kind of 
creatures as our parents. They 
gave rise to us, other humans — not to a 
plant, or a fish, or a bird. Let us start, 
therefore, by assuming the existence of some 
intrinsic factor which determines that hu- 
mans shall beget humans, and let us call this 
inborn factor for the genesis of like from 
like the genetic factor. Since each plant and 
animal produces offspring of its own kind, 
or species, we can generalize and hypothesize 
that every species of organism has such a 
built-in genetic factor. But it must also be 
admitted that the genetic factors for dog, 
for apple tree, and for man all differ in some 
way in order to produce such different or- 
ganisms as end products. 

In addition to basic likenesses within the 
species, each person is similar to and differ- 
ent from his parents in respect to certain 
details. What is the basis for this? If par- 
ents and offspring have similar caloric in- 
takes, all will weigh more nearly alike at a 
comparable age than if their caloric intakes 
were different. Apparently, then, environ- 
ment in which parents and children live can 
sometimes be the cause of their similarities 
and differences. But are all similarities and 
differences among human beings produced 
by environment, or does the genetic factor 
presumed responsible for like begetting like 
play a role in their production? 
1 



In trying to formulate the answer to this 
question, it may be helpful to consider the 
results of certain studies with bean plants. 1 
The particular kind of bean plant concerned 
reproduces sexually, a single plant perform- 
ing the functions both of male and female 
parent. For the present, assume that the 
genetic ) actor is transmitted from the parent 
to the offspring, and that the transmitted 
factor must be the same as that of the par- 
ent. Assume also that the genetic factor 
has a natural rather than a supernatural or 
spiritual basis. If the genetic factor has a 
natural basis then it ought to have a material 
basis and have chemical and or physical 
properties, as have other material things. 
One is led, therefore, to postulate the exist- 
ence of genetic material. 

Genetic Material 

Consider a particular bean. When the plant 
grown from this seed produces offspring 
beans (Figure 1-1 A), these are found to 
vary in size, some being very small, some 
small, and some medium. According to the 
assumptions made, these beans all have the 
same type of genetic material or genetic con- 
stitution — or genotype. The simplest ex- 
planation one can offer for the size differ- 
ences is that they were caused by environ- 
mental differences occurring during seed for- 
mation. This view can be tested by growing 
each of the beans and scoring the size of 
seeds each produces. When this is done, 
each bean is also found to produce offspring 
beans of very small, small, and medium 
sizes, regardless of the size of the parent 
seed itself. This test can be repeated gen- 
eration after generation with the same result. 
Such a line of descent, whose members carry 
the same genotype, can be called a pure line. 
The manifestation of the genotype in traits 
or characteristics (size, in our example) is 

1 Based upon W. Johannsen's experiments. See 
reference on p. 12. 



CHAPTER 1 



Typical 
Offspring 

Typical 
Offspring 



/l\ /l\ /l\ 



PURE LINE I 



B 



/l\ /l\ 



PURE LINE II 



/l\ /l\ /i\ 



* 




• • • ^ 



NEW MUTANT 
PURE LINE 



OLD PURE LINE 

figure 1-1. Relative sizes of seeds obtained from self-fertilized bean plants. 



Genetic Material and Mitosis 



called the phenotype. Environmental differ- 
ences can cause the same genotype to pro- 
duce a variety of phenotypes, and one can 
conclude that the differences between the 
beans of a pure line are environmentally pro- 
duced and are not due to differences in 
genotype. 

Now consider another bean, of the same 
species, which gives rise to offspring beans 
(Figure 1-1 B) that are very large, large, 
and medium sized. Since each of these pro- 
duces offspring beans which again show the 
same range of phenotypes. another and dif- 
ferent pure line is clearly involved, within 
which phenotypic variability is attributable 
to environmental fluctuation. 

How can one explain the differences be- 
tween these two different pure lines, one 
producing some very small and small beans 
and the other producing some very large and 
large ones? Since all the beans are grown 
under the same environmental conditions, 
these phenotypic differences cannot be due to 
environmental differences; instead they must 
be due to genotypic differences. It must be 
concluded, then, that the genetic material in 
these two pure lines is different. How can 
one explain that some of the seeds in both 
of these genotypically different pure lines 
are similar — medium sized? Apparently, 
different genotypes have produced the same 
phenotype due to the influence of the en- 
vironment. 

As already mentioned, under similar en- 
vironmental conditions the average size of 
the beans produced within a pure line re- 
mains the same regardless of the size of the 
specific beans planted. That is, in the pure 
line first described the offspring beans have 
the same average size whether the very small 
or the medium seed is used as parent. Sim- 
ilarly, the average size of seed produced in 
the second pure line is the same whether the 
medium or the very large seed is the parent. 
In other words, selection within pure lines 



is futile, as expected in view of the hypoth- 
esis that all members of a pure line are ge- 
netically identical. 

Throughout the bean experiments de- 
scribed, every effort was made to keep the 
environment the same. This does not mean 
that the environment did not vary, but that 
it varied approximately in the same ways 
and to the same extent for all the groups in 
the study. In this particular work it happens 
that phenotypic variability due to the fluc- 
tuations of environment is not so great as to 
mask completely the phenotypic effect of a 
genetic difference. In any randomly chosen 
case, however, one cannot predict offhand to 
what degree any particular phenotype will be 
influenced by the genotype and by the en- 
vironment. Hypothetically, then, two indi- 
viduals of the same species can have both 
phenotypic similarities and phenotypic dif- 
ferences resulting from each of the following 
four combinations, as the examples indicate: 

1. Identical genotypes 

in near-identical environments 
Phenotypic difference — one small and one 

medium sized bean from the same pure 

line. 
Phenotypic similarity — two small sized 

beans from the same pure line. 

2. Different genotypes 

IN NEAR-IDENTICAL ENVIRONMENTS 

Phenotypic difference — one small and one 

large bean from genetically different 

pure lines. 
Phenotypic similarity — two medium sized 

beans from genetically different pure 

lines. 

3. Identical genotypes 

in different environments 
Phenotypic difference — one bean plant 
grown in the light is green, while an- 
other grown in the dark is white, 
though both are from the same pure 
line. 



( II \l' I I K 1 




FIGURE 1-2. Male Siamese cat, grown under 
temperate conditions, showing the same pig- 
mentation pattern as the Himalayan rabbit. 
{After C. E. Keeler and V. Cobb.) 



Phenotypic similarity — two rabbits from 
a certain pure line (genetically black 
rabbits) both have black coats even 
though one individual grew at high 
temperatures and the other grew at 
low temperatures. 
4. Different genotypes 

in different environments 

Phenotypic difference — a rabbit from a 
genetically black line, grown in a cold 
climate, has black fur, while a rabbit 
from a Himalayan line, grown under 
temperate conditions, is Himalayan, 
i.e., white except for the extremities 
(paws, tail, snout, and ears), which are 
black (see Figure 1-2). 

Phenotypic similarity — a rabbit from a 
genetically black line grown at a mod- 
erate temperature and a rabbit from a 
genetically Himalayan line grown at a 
cold temperature both have black fur. 

The final example illustrates that genotypi- 
cally different individuals which are pheno- 
typically different in one environment may 
become phenotypically similar when placed 
in different environments. The all-black 
Himalayan rabbit is termed a phenocopy of 
the genetically black rabbit. Persons who 



arc genetically diabetic and take insulin arc 
phenocopies of genetically normal persons 
who do not take insulin. Genetically normal 
embryos whose mothers are exposed to the 
drug thalidomide develop into phenocopies 
o\' genetically abnormal, phocomelic persons 
lacking most or all of the lour limbs. So 
both normal and abnormal phenotypes can 
be phenocopied. 

The case of coat color in rabbits is instruc- 
tive in another respect. The rabbit that is 
genetically black will always produce a black 
coat no matter what the temperature is, pro- 
vided the temperature is not lethal. In the 
case of this genotype there seems to be no 
range of phenotypic expression with respect 
to temperature variations. In the Himalayan 
strain, however, the situation is different, as 
already described in part. If grown at very 
high temperatures such rabbits have entirely 
white coats. In this case the phenotypic 
range of reaction, or norm of reaction, of 
the genotype is relatively great, varying with 
increasing temperature from completely 
black through the Himalayan pattern to 
completely white. 

We are now in a position to answer the 
question concerned with the basis of simi- 
larities and differences among offspring or 
between them and their parents. Extending 
the principles just described for beans and 
rabbits to all other kinds of organisms, in- 
cluding man, it is concluded that not only is 
the genetic material different in different 
species of organisms, but that it can also 
differ from one organism to another in the 
same species. Phenotypic similarities be- 
tween individuals may occur when they are 
carrying the same or different genotypes, and 
phenotypic differences between individuals 
may or may not be accompanied by geno- 
typic differences. 

Having agreed that genetic variation 
exists within as well as between species, 
one may now ask: How does genetic varia- 



Genetic Material and Mitosis 



tion arise? If a pure line of large beans is 
bred for many generations, one finds, on rare 
occasions, a very small bean which gives 
rise to offspring beans ranging from tiny to 
small, and which clearly make up a new. 
different, pure line (Figure 1-1C). What 
has apparently happened is that the genetic 
material in the pure line of large beans 
somehow changed to another transmissible 
form which henceforth caused the produc- 
tion of beans which are, on the average, very 
small. Such a change in the genotype that 
is transmitted to progeny may be attributed 
to a process called mutation. The result of 
mutation is a mutant, a term which is ap- 
plicable to either or both the genotype and 
the phenotype of the new kind of individual. 

Just as it is easy to ascribe differences be- 
tween dogs and cats to genetic differences, 
so it is often simple to tell that certain dif- 
ferences between lines of the same species 
have a genetic basis. There are many strains 
or breeds of pigeons, dogs, cattle, and of 
other domesticated animals, each of which 
differs from the other in phenotype. That 
many of these differences are due to genetic 
differences is proved by finding that the 
phenotypic differences are retained even after 
the different breeds are raised together gen- 
eration after generation in essentially iden- 
tical environments. Revealed in this way, 
the genotypes within a species are of im- 
mense variety. This already-present genetic 
variation should be kept in mind in seeking 
to learn more about the nature of the genetic 
material. 

In order to learn more about the genetic 
material, the material things comprising or- 
ganisms can be examined more closely, par- 
ticularly those substances transmitted from 
parent to offspring. Most types of organ- 
isms are composed of (usually microscopic) 
building blocks, or cells, plus substances that 
have been manufactured by cells. Such an 
organism begins life either as a single cell, 



or by the fusion of two cells into one, or as 
a group of nonfusing cells derived from the 
parents. The cell serves as the link or bridge 
between generations. In those cases where 
the new individual begins life as one cell or 
as a group of nonfusing cells derived from 
a single parent, reproduction is asexual, 
whereas in cases where two parents con- 
tribute cells, reproduction is sexual. In sex- 
ual reproduction two mature sex cells, or 
gametes, fuse in the process of fertilization 
into one new cell, the zygote, which is the 
start of a new individual. In higher animals 
the gametes are called egg (female) and 
sperm (male), and the zygote the fertilized 
egg. In the bean plant, as already men- 
tioned, male and female gametes are pro- 
duced in the same individual and self-fertili- 
zation normally occurs; in human beings the 
two kinds of sex cells are produced in sepa- 
rate individuals of different sex, so that cross- 
fertilization always occurs. 

When might the hypothesized genetic ma- 
terial be transferred from parent to offspring? 
Consider certain organisms, composed of 
only a single cell, which reproduce asexually 
by dividing into two cells. In this process 
the parent becomes extinct, so to speak, its 
individuality being replaced by two daughter 
cells of the same kind. Once formed, the 
two daughters often separate, never to meet 
again. In such a case, the genetic material 
must have been transmitted before the com- 
pletion of cell division. Accordingly, the 
cell and this process of cell division should 
be studied in some detail for clues concern- 
ing the physical basis and transmissive char- 
acteristics of the genetic factors. 

Mitosis 

Attention has already been called to the 
cellular bridge between generations. It is 
only via this bridge that genetic transmission 
may take place, at least in single-celled or- 
ganisms for which cell division is equivalent 



6 



CHAPTER 1 



to reproduction. All cellular organisms are 
remarkably similar in the way they accom- 
plish cell division. Accordingly, let us ex- 
amine briefly certain general features of cell 
structure and the appearance, under the 
microscope, of cells undergoing division, in 
initiating our search for the material basis 
of the genotype. 

There are two major parts of the cell 
(Figure 1-3): a peripheral portion compris- 
ing the cytosome, containing substances mak- 
ing up cytoplasm, and a more central portion 
called the nucleus, containing nucleoplasm. 
In the final stages of cell division in higher 
plants, the cytoplasm is divided by a cell 
plate, whose growth starts internally and 
proceeds toward the periphery until the sep- 
aration into two daughter cells is complete. 
In the case of animal cells, a furrow starts 
at the periphery of the cell and deepens until 
the parent cell is cleaved into two. The 
degree to which the two daughter cells are 
identical with respect to cytoplasmic com- 
ponents depends upon the position of the 
cell plate or furrow in the parent cell. In 
some cases these occur in the middle of the 
cell, but in many other cases they are located 
off-center, producing daughter cells which 
contain very different amounts of cytoplasm. 
Although the cytoplasmic components of a 
parent cell are often distributed unequally 
between daughter cells, this is not true for 
the nuclear contents. Ordinarily, nuclear 
division directly precedes cytosomal division. 
But the nucleus does not simply separate 
into two parts by the formation of a furrow 
or cell plate. Instead, the nucleus under- 
goes a remarkable series of activities in order 
to divide; this whole process of indirect nu- 
clear division is called mitosis. 

During the time that a nucleus shows no 
visible evidence of mitosis, it is nevertheless 
very active biochemically. In appearance 
(Figure 1-4A), it is bounded by a nuclear 
membrane and is filled by a more or less 



homogeneous-appearing ground substance or 
matrix in which are located one or more 
small bodies, called nucleoli. 

The first indication that the nucleus is pre- 
paring to divide is the appearance in its 
ground substance of a mass of separate fibers 
(Figure 1-4B), some of which seem to be 
associated with the nucleoli. These fibers 
are called chromosomes, and their appear- 
ance marks the start of the first phase of 
mitosis, or prophase. Careful cytological 
observation reveals that each chromosome 
is in turn composed of two major delicate 
threads irregularly coiled about each other. 
Each of the paired threads within each chro- 
mosome is called a chromatid. As prophase 
continues, the chromatids within each chro- 
mosome become shorter and thicker and 
untwist from each other (Figure 1-4C). 
Some of the material incorporated to thicken 
the chromatids may be derived from the 
nucleoli, which are seen to become smaller. 
By the end of prophase (Figure 1-4D), the 
nucleoli and nuclear membrane have disap- 
peared and the chromatids have formed 
thick rods which begin to move actively for 
the first time. Active motility is not the 
property of the entire chromosome, however, 
but is restricted to a particular region of it 
called the centromere or kinetochore (see 
p. 379). 

The centromeres move in a particular di- 
rection relative to a fibrillar structure called 
the spindle which has been forming through- 
out prophase. The completed spindle has 
a shape similar to what is produced when 
one extends and separates the fingers and 
touches corresponding fingertips together. 
The wrists represent the poles of the spindle 
and the fingers, the spindle fibers. The chro- 
mosomes migrate from whatever position in 
the spindle region they may have, until each 
centromere comes to lie in a single plane 
perpendicular to the axis between the poles, 
that is, at the equatorial plane or equator 



Genetic Material and Mitosis 




n>T© 1 7967 D A 'v7""" f f " "' SeC " 0n "' " "' L ^0"™« »'<"< amission. Copy, 
right © 1961 by Scientific American, Inc. All rights reserved.) 



s 



CHAPTER 1 



of the spindle, which is represented by the 
plane formed where the fingertips touch. 
The rest of eaeli chromosome, being pas- 
six e. can be in an\ position in the spindle. 
When all the centromeres have arrived at 
the equatorial plane of the spindle, mitosis 
has reached the middle phase, or metaphase 
(Figure 1-4E). 

Until this point the chromatids of a chro- 
mosome are still attached to each other at or 
near the centromere, although elsewhere they 
are largely free. Next they also separate 
at the centromere and the two daughter 
centromeres suddenly move apart, one going 
toward one pole of the spindle, the other 
toward the other pole, with the rest of each 
chromatid, which is now recognized as a 
chromosome, being passively dragged along. 
This stage, in which the chromatids separate, 
move toward, and arrive at the poles as chro- 
mosomes, is called anaphase (Figure 1-4F). 

When the chromosomes have reached the 
poles, the last stage, or telophase, occurs 
(Figure 1-4G), in which the events appear 
to be the reverse of those that happened in 
prophase. Specifically, the spindle disinte- 
grates, a new nuclear membrane is formed 
around the chromosomes, and nucleoli re- 
appear. The chromosomes become thinner 
and longer and then can be seen to consist 
of two delicate threads (chromatids) wound 
one about the other. Finally, as the chromo- 
somes lose their visible identity, the nucleus 
enters the interphase, inter mitotic , or meta- 
bolic stage (see again Figure 1-4A). 

The impression may have been gained 
that, in one respect, the preceding general- 
ized account of the mitotic phases was either 
incomplete or misleading. It was stated that 
the prophase chromosome is composed of 
two chromatids or threads, that metaphase 
puts these into position for separation at 
anaphase, and that after separation their 
newly attained individuality is recognized by 
calling them chromosomes. But chromo- 



somes were defined as containing two visible 
threads! The question rightly asked is: does 
the anaphase chromosome contain the two 
threads that are later seen at telophase? 
This would be true if each chromatid some- 
how visibly reproduced itself between the 
time it was seen relatively uncoiled at pro- 
phase and the next time it was seen rela- 
tively uncoiled, at telophase. Remember 
that we have been discussing the replication 
of chromatids as detected by microscopic 
observation. Chromosome and chromatid 
replication can also be studied by other 
means. Let us consider some evidence re- 
garding chromosome replication at the chem- 
ical level, which may help us understand its 
replication at the visible level. 

Chromosomes ("colored bodies") are 
unique since they are the only objects in 
the cell that are made purple by the Feulgen 
staining technique. It is possible to measure 
the amount of chromosomal material by the 
amount of purple stain held by the chromo- 
somes. The amount of chromatin — Feul- 
gen-stainable chromosomal material — does 
not change between prophase and telo- 
phase, but doubles over a period of hours 
during the intermitotic stage. By the be- 
ginning of prophase, therefore, each chro- 
mosome, as revealed by its stainability, 
has already replicated chemically. At the 
visible level, however, this is not yet ap- 
parent, so that each of the two visible 
chromatids in a chromosome also contains 
the chemical materials for an identical chro- 
matid which is still not resolved as a sepa- 
rate thread under the microscope. This new 
material is unresolved either because it has 
not yet assumed a proper chromatid form 
or has done so but is so tightly paired with 
its sister chromatid that together they ap- 
pear as one strand. Before the next occa- 
sion when unwound threads can be seen — 
that is, at the telophase of the same mitosis 
— this replication at the visible level has al- 



Genetic Material and Mitosis 




10 



CHAPTER 1 



read) been accomplished. Thus, the chem- 
ical replication that takes place in a given 
interphase is not visible in chromatid form 
until the succeeding telophase. 

What are the consequences of mitosis'.' 
Speaking in terms of visible structures, the 
chromosomal content of the parent nucleus 
has become repeated in each daughter nu- 
cleus, so that the subsequent division of the 
cytoplasm produces daughter cells whose 
chromosomal composition is identical to each 
other's and to that of the parent cell from 
which they were derived. Mitosis merely 
provides the cellular machinery for the exact 
partitioning of previously replicated chromo- 
somal material. The cells of different species 
are different in that they have different num- 
bers of chromosomes per nucleus, or the 
chromosomes differ in appearance, or both. 
One chromosome may differ from another 
in size, in stainability with various dyes, and 
in the position of the centromere. Most 
chromosomes have a single centromere which 
is not located terminally, i.e., at an end, and 
therefore separates the two arms of a chro- 
mosome; all chromosomes and chromatids 
are unbranched fibers. 

Examination of the kinds of chromosomes 
present at metaphase in sexually reproducing 
organisms typically reveals that for each 
chromosome which arrives at the equatorial 
plane, there is another chromosome very 
similar or identical in appearance which also 
takes a position independently in this plane. 
Chromosomes thus occur as pairs; the mem- 
bers of a pair are called homologous chro- 
mosomes, or homologs, whereas chromo- 
somes of different pairs are nonhomologous, 
or nonhomologs. It should be repeated that 
the members of a pair of homologs take 
their positions at mitotic metaphase inde- 
pendently of each other. 

The number of chromosomes seen in 
typical mitosis of the garden pea is 7 pairs; 
in Indian corn (maize) there are 10 pairs, 



in the domesticated silkworm 28, and in 
human beings 23; thus the chromosomes of 
a species are characteristic in number - as 
well as in form. Whatever the number o\ 
chromosomes present in the zygote, then, 
other things being equal, the same number 
of chromosomes will be found in every cell 
of a multicellular organism descended from 
the zygote by cell divisions in which mitosis 
has occurred. 

Chromosomes as Genetic Material 

The chromosomes are one of the character- 
istic components transmitted by all cells to 
daughter cells. Chromosomes reproduce 
themselves and are transmitted in mitosis 
equally to the daughter cells so that these 
are identical, in this respect, to each other 
and to their parent cell. Let us make the 
additional reasonable assumptions that ge- 
netic material arises only by the replication 
of pre-existing genetic material, and also that 
different genotypes arise only from each 
other by mutation, that is, by a genotype's 
changing to an alternative mutant form which 
in turn is involved in reproducing the alter- 
native form until it undergoes mutation. A 
chromosome may occasionally become visi- 
bly altered in certain ways; in these cases all 
chromosomes ordinarily derived from such a 
modified chromosome via mitosis have ex- 
actly the same alteration. Therefore, both 
genetic material and chromosomes are con- 
sidered capable of mutation and are subse- 
quently involved in replicating their new 
form. On this basis, then, we can hypoth- 
esize that the chromosome is, or carries, the 
genetic material. 

It has been implied that the genetic ma- 
terial routinely retains its individuality or 
integrity regardless of the nature of the en- 
vironment. One indirect piece of evidence 
has already been cited for believing this is 

-'See S. Makino (1951) and C. D. Darlington and 
E. K. Janaki-Ammal (1945). 



Genetic Material and Mitosis 



11 



true for the chromosomes — namely the in- 
dependence with which each chromosome 
arrives at metaphase. It might be thought, 
when the chromosomes "disappear" during 
interphase, that their individuality is lost 
and even that their contents are dispersed. 
That the nuclear material is not dispersed 
into the cytoplasm between successive mi- 
toses is indicated by the retention of the 
full amount of chromosomal material within 
the nucleus during interphase, insofar as re- 
vealed by the Feulgen staining technique. 
Even so, it is still possible that those com- 
ponents of chromosomes which remain in 
the nucleus become scrambled during inter- 
phase and later resynthesize their proper 
form during the next prophase. Four lines 
of evidence bearing on this matter can be 
mentioned. The first three come from study- 
ing the appearance of successive mitoses. It 
is possible to observe the relative positions 
of the chromosomes at late anaphase or telo- 
phase and also their relative positions as 
they enter the next prophase. When this is 
done, the chromosomes are found to have 
held the same relative positions, as expected 
had they retained their integrity during the 
intervening interphase. Second, since the 
nucleolus does not fragment during inter- 
phase, those parts of the chromosomes, 
called nucleolus organizers, with which the 
nucleolus is associated probably remain as- 
sociated during that interval. 3 Third, it 
sometimes happens that two originally iden- 
tical homologs are modified by mutation so 
that each is changed in a different respect. 
The finding, mitosis after mitosis, that both 
homologs retain their separate differences is 
evidence that each homolog has retained its 
individuality cell generation after cell gen- 
eration. Finally, more direct evidence on 
the retention of chromosomal individuality 
during interphase is available from cells of 

3 See also F. H. Ruddle (1962). 



larval salivary glands of certain flies. These 
giant cells have interphase nuclei that con- 
tain giant chromosomes which, though rela- 
tively uncoiled, are clearly equivalent to the 
more contracted chromosomes seen during 
mitosis. 

The number of points of similarity be- 
tween genetic material and chromosomes is 
already impressive. However, if all nuclei 
divide by mitosis, a gamete should contain 
the same number of chromosomes as the 
other cells derived from the original zygote; 
and since the zygote of any generation com- 
bines two gametes, the number of chromo- 
somes should increase in the zygotes of suc- 
cessive generations. One would therefore 
expect an increase in the amount of genetic 
material in successive sexual generations. 
This expectation is not realized, however, 
since one finds that all individuals of a 
species have a characteristic, typically stable, 
chromosomal content. In fact, as expected, 
human gametes do not contain the paired, 
diploid, chromosome number, that is, 23 
pairs of nonhomologous chromosomes. In- 
stead each usually contains 23 chromosomes, 
one of each nonhomologous type, comprising 
a complete, unpaired, haploid or monoploid, 
set of chromosomes. The zygote, therefore, 
has the diploid chromosome constitution re- 
stored because each gamete furnishes a 
haploid set of chromosomes, one set con- 
tributed by the sperm from the father, and 
another set by the egg from the mother. In 
this way chromosomes remain as pairs, 
sexual generation after sexual generation, 
and the number of chromosomes in zygotes 
remains unchanged. Clearly, then, the cell 
divisions preceding gamete formation cannot 
be invariably mitotic, but must involve at 
some point a special mechanism for reducing 
the number of chromosomes from diploid to 
haploid. The nature of this special kind of 
nuclear behavior is considered in the next 
chapter. 



12 CHAPTER 1 

SUMMARY AND CONCLUSIONS 

Organisms are assumed to contain an intrinsic genetic factor which is responsible for 
like reproducing like. Ibis genetic factor is presumed to have a physical basis in genetic 
material. 

The genetic material must be different in dilferent species of organisms, and may 
be different in different lines or breeds of the same species. Variations in phenotype 
m.i\ be due to genetic or environmental differences, or both. The contribution made 
to phenotypic variability bj one of these two factors may be evaluated by holding the 
other factor as constant as possible. 

Genotypic differences arise by the process of mutation. The genetic material is 
presumably transmitted from parents to offspring by means of the cellular bridge be- 
tween generations, and is assumed to be self-replicating and to arise only from pre- 
existing genetic material. 

Studies of cell division in which nuclei divide mitotically reveal that, of all cellular 
components, the chromosome is the structure most likely to serve as the genetic material 
or as its carrier. This hypothesis receives support from several of the properties of 
chromosomes which parallel established or assumed properties of genetic material. 
Chromosomes come only from pre-existing chromosomes; different species have differ- 
ent chromosomal compositions; the chromosome content is identical both quantitatively 
and qualitatively in each cell of a line produced by asexual reproduction; each chromo- 
some retains its individuality, mitotic cell generation after mitotic cell generation, re- 
gardless of the nature of the other chromosomes present; chromosomes can occasionally 
mutate, the mutant chromosome then replicating the mutant form. 



REFERENCES 

Darlington, C. D., and Janaki-Ammal, E. K., Chromosome Atlas of Cultivated Plants, 
London: Allen and Unwin, 1945. 

Flemming. W., 1 879. "Contributions to the Knowledge of the Cell and its Life Phe- 
nomena," as abridged and translated in Great Experiments in Biology, Gabriel, 
M. L., and Fogel, S. (Eds.), Englewood Cliffs, N.J.: Prentice-Hall, 1955, pp. 
240-245. 

Johannsen, W., 1909. Elemente der exakten Erblichkeitslehre. Jena. See also a trans- 
lation of the summary and conclusions of his 1903 paper, "Heredity in Populations 
and Pure Lines," in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood 
Cliffs, N.J.: Prentice-Hall, 1959, pp. 20-26. 

Makino, S., An Atlas of Chromosome Numbers in Animals, Ames, Iowa: Iowa State 
College Press, 1951. 

Mazia, D., "Mitosis and the Physiology of Cell Division," in The Cell, Vol. 3, Meiosis 
and Mitosis, pp. 77-412, Brachet, J., and Mirsky, A. E. (Eds.), New York: Aca- 
demic Press, 1961 . 

Ruddle, F. H., "Nuclear Bleb: A Stable Interphase Marker in Established Lines of 
Cells in Vitro," J. Nat. Cancer Inst., 28:1247-1251, 1962. 

Schrader, F., Mitosis: the Movement of Chromosomes in Cell Division, New York: 
Columbia University Press, 1953. 

Scientific American, Sept. 1961, Vol. 205, No. 3, "The Living Cell," articles by J. 
Brachet and D. Mazia. 



Genetic Material and Mitosis 13 




WlLHELM LUDWIG JOHANNSEN 

(1861-1926). (From Genetics, 
vol. 8, p. 1, 1923.) 



Spector, W. S. (Ed.), "Chromosome Numbers," in Handbook of Biological Data, Phila- 
delphia: Saunders, 1956, pp. 92-96. 

Swanson. C. P., Cytology and Cytogenetics, Englewood Cliffs, N.J.: Prentice-Hall, 1957. 

Swanson, C. P.. The Cell, 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1964. 

QUESTIONS FOR DISCUSSION 

1.1. Does the phenotype of one generation have any effect upon the genotype of the 
next? Explain. 

1.2. Evaluate the thesis that the genotype is more important to organisms than is the 
environment. 

1.3. Is the environment for two organisms ever identical? Explain. 

1.4. What is meant by an operational definition? 

1.5. Define the genetic factor. Have you given an operational or a nonoperational 
definition? Explain. 

1.6. When the same similarities or differences in phenotype can be produced by either 
the environment or the genotype, can one ever be sure which is the determining 
factor? Explain. 

1.7. What evidence can you give to support the view that genetic material is trans- 
mitted from parent to offspring? Do you think this evidence constitutes conclu- 
sive proof of transmission? Explain. 

1.8. What conclusions can you reach regarding the genetic factor in Himalayan rabbits 
and in Siamese cats? 



14 CHAPTER 1 

1.9. Assume the genetic factor has a supernatural basis. Could we learn anything 
about it b> the scientific method of investigation? Explain. 

l.K). Do you think human beings provide good material for the study of the genetic 
factor? Explain. 

1. 11. What si/e limitations can you give to the genetic material? 

1.12. Is the existence o[ genetic material presumed or proved? Explain. 
LI 3. What do the bean experiments reveal about genetic material? 

1.14. August Weismann ( 1834-1914) cut off the tails of mice for a series of genera- 
tions and found that tail length remained normal each new generation. Why 
are these experiments significant? 

1.15. What are the consequences of mitosis? 

1.16. For each of the properties of chromosomes listed in the Summary and Conclu- 
sions, state the corresponding property of the genetic material and identify it as 
one that is either proved or assumed. 

1.17. [f the chromosomes serve as the genetic material, each cell of the body derived 
by mitosis should carry the same genotype. Describe how you would test this 
idea, using a multicellular plant. 

1.18. What are the advantages or disadvantages of chromosome coiling? 

1.19. Can you imagine a spindle which is too small for normal cell division? Explain. 

1.20. Suppose certain nuclei normally do not divide with the aid of a spindle. How 
would this affect your ideas about genetic material? 

1.21. Discuss the statement that all cell divisions are normally mitotic. 

1.22. Differentiate between replication of chromatids and of chromosomal material. 

1.23. List the events that presumably take place before a given telophase chromosome 
can give rise to a chromosome made entirely of chromosomal material not yet 
synthesized. 

1 .24. Why should the peas in a pod be similar? Different? 

1.25. What do each of the following observations mean with regard to the origin and/or 
integrity of chromosomal material? 

(a) Nonhomologous chromosomes retain their characteristic morphological dif- 
ferences mitosis after mitosis. 

(b) A loss or gain of one entire chromosome occurs occasionally, with all mitotic 
descendants having the same aberration. 

(c) T. Boveri noted in Ascaris cleavage that sister cells entering the next mitosis 
often have a mirror-image arrangement of their chromosomes. 

1.26. What conclusions can you draw from the fact that there are three genotypically 
different kinds of Indian corn: one always has red kernels, one always has yellow 
kernels, and one has kernels which are yellow but become red if exposed to 
sunlight? 



Chapter 2 

MEIOSIS AND 
CHROMOSOMAL SEGREGATION 



H 



"ow do both male and female 
gametes come to contain 
only one set of chromo- 
somes, composed of one member of each 
pair of chromosomes found in the nucleus 
of an ordinary body, or somatic, cell? If 
gametes were produced by regular mitotic 
division, they would be diploid. The reduc- 
tion from two sets to one is brought about 
by another type of indirect nuclear process, 
called meiosis, which actually requires two 
successive nuclear divisions to accomplish 
its result. 

Meiosis 

To render the cytological description of the 
meiotic process more meaningful, several as- 
sumptions will be made. Suppose that the 
processes directing the division of the nucleus 
act especially early in the case of meiosis, 
before the chromosomes have attained the 
degree of coiling first seen in mitotic pro- 
phase. Suppose further that a relatively 
more uncoiled state of the chromosome is, 
under these conditions, associated with an 
especially strong attraction between homo- 
logs of like chromosome parts for like parts 
and that this attractive force extends over 
considerable, though still microscopic, dis- 
tances. Then, with one additional novelty 
yet to be described, the meiotic process will 
occur in the following predictable way when 
the chromosomes, without further replica- 
15 



tion, undergo two successive mitotic divi- 
sions. 

In prophase of the first meiotic division, 
just as in mitotic prophase, each chromo- 
some contains two chromatids plus an equal 
amount of chromosomal material not yet 
visible as chromatids (see p. 8). But 
now, because of the early onset of nuclear 
division, homologous chromosomes pair 
point for corresponding point (making a 
bundle of four chromatids plus an equal 
amount of future chromatid material). Ac- 
cordingly, the chromosomes proceed as pairs 
to the equator of the spindle for the meta- 
phase. (Recall that in mitosis, on the other 
hand, each chromosome of the two sets pres- 
ent goes to the equator of the spindle in- 
dependently of its homologous chromo- 
some.) At anaphase the members of a pair 
separate and go to opposite poles, each ana- 
phase chromosome still containing two chro- 
matids plus an equivalent amount of future 
chromatid material. In the interphase after 
the first telophase, no synthesis of future 
chromatid material takes place since what 
was made in the previous interphase had 
not been used to make visible chromatids 
in the first meiotic division. The second 
meiotic division may start at any time and 
proceed as a typical mitosis. In the second 
meiotic prophase each chromosome contains 
two chromatids and the material for two 
future chromatids. Each chromosome pro- 
ceeds to metaphase independently; at ana- 
phase the two chromatids separate and go 
to opposite poles of the spindle (after sep- 
aration the chromatids may be called chro- 
mosomes). By telophase the future chro- 
matid becomes visible; thus each telophase 
chromosome contains two chromatids. 

Although mitosis always involves chromo- 
some duplication and separation alternately, 
one duplication is followed by two separa- 
tions in meiosis. The result is the mainte- 
nance of the diploid chromosome condition 



IT) 



CHAPTER 2 



in mitosis, but a reduction from the diploid 
to the haploid condition upon the completion 
of meiosis. 

I et us examine in some detail the actual 
meiotic process as seen under the micro- 
scope (Figure 2-1 ). Prophase of the first 
meiotic division (prophase I ) is of long dura- 
tion, as compared to mitotic prophase, and 
is divided into several substages, each with 
its own distinguishing characteristics. 

1 . As they emerge from the interdivision 
phase the chromosomes are long and thin, 
more so than in the earliest prophase of 
mitosis. This is the leptonema (thin thread) 
stage of prophase I. 

2. Next the thin threads pair with each 
other in a process called synapsis. This 
pairing is very exact, being not merely 
between homologous chromosomes, but be- 
tween exactly corresponding individual 
points of the homologs. 1 Synapsis proceeds 
zipperwise until the two homologs are com- 
pletely apposed. This is the zygonema (join- 
ing thread) stage. 

3. The apposition of homologs becomes 
so tight that it is difficult to identify two sep- 
arate chromosomes in the pachynema (thick 
thread) stage (Figure 2-2A). 

4. Next, the tight pairing of the pachy- 
nema is relaxed, whereupon it can be clearly 
seen in the diplonema (double thread) stage 
that each pair of synapsed chromosomes 
contains four threads, two visible chromatids 
for each chromosome (Figure 2-2B, C). A 
pair of synapsed chromosomes is called a 
bivalent (composed of two univalents) when 
referring to chromosomes, but is called a 
tetrad (composed of two dyads or four 
monads) when referring to cytologically de- 
tectable chromatids. 

Although the chromatids in a tetrad sepa- 
rate from each other in pairs here and there, 
they are all still in close contact with each 
other elsewhere. Each place where the four 

1 See H. Jehle (1963) for a discussion of the 
physical basis for the attraction of like for like. 




figure 2-1 . Meiosis in the lily 
(Courtesy of R. E. Cleland.) 



-general view. 



chromatids are still held together is called 
a chiasma (Greek, cross; plural, chiasmata) 
(Figure 2-3 A). In a chiasma the two chro- 
matids that synapse to make a pair on one 
side of the point of contact, separate at that 
point and synapse with other partners on 
the other side of the contact point; i.e., the 
partners making up two synapsed pairs of 
chromatids are different on the two sides of 
the place of contact (Figure 2-3B). A 
tetrad typically has at least one chiasma. 
The occurrence of a chiasma assures that 
the univalents are held together. When sev- 
eral chiasmata occur per bivalent, loops are 

figure 2-2 (opposite) . Meiosis in the lily. 
The leptonema and zygonema stages of pro- 
phase I have been omitted. (Courtesy of R. E. 
Cleland.) (By permission of McGraw-Hill 
Book Co., Inc., from Study Guide and Work- 
book for Genetics, by I. H. Herskowit: :. copy- 
right I960.) 



Meiosis and Chromosomal Segregation 



17 




A. PACHYNEMA 



DIPLONEMA 



D. DIAKINESIS 



PROPHASE I 



T H 




4ifi&Jv 



I , 




L. 




${ 'te^ 




E. METAPHASE I F. ANAPHASE I G. INTERPHASE I 

(Equatorial View) (Side View) 




■Mwr 







H. METAPHASE II 
(Side View) 



I. TELOPHASE II 



is 



CHAPTER 2 



formed, adjacent ones at right angles to each 
other. 

As diplonema continues, the chromosomes 
become shorter and thicker, more compacted 
than they ever become in mitosis. 

5. In some animals, during the formation 
oi female gametes especially, a diffuse or 
growth stage follows diplonema, in which 
the chromosomes ami nucleus revert to the 
appearance kmnd in a nondividing cell. 
During this stage a great amount of cyto- 
plasmic growth takes place. In human be- 
ings this stage may last for decades, after 
which the rest of meiosis occurs and mature 
eggs ready for ovulation are produced. 

6. Diakinesis (Figure 2-2D) is charac- 
terized by the maximal contraction of diplo- 
nema chromosomes, or by maximal recon- 
traction of the chromosomes which had 
entered a diffuse stage. By the end of this 
stage nucleoli and nuclear membrane have 
disappeared, the spindle has formed, and 
prophase I is completed. 

Metapha.se I (Figure 2-2E) is attained 
by the movement of chromosomes to the 
midspindle, as in mitosis, except that they 
move as bivalents, made up of a tetrad of 
chromatids still held together by chiasmata. 
Between diplonema and metaphase I the 
chiasmata move toward the end of the chro- 
mosome arms, that is. away from the centro- 
mere, especially if the bivalent is short. As 
a consequence of this chiasma terminaliza- 
tion the number of chiasmata present at 
metaphase I may be less than it was at diplo- 
nema. 

During anaphase I (Figure 2-2F) the uni- 
valents in each bivalent separate from each 
other at the region of the centromere and 
proceed to opposite poles of the spindle. 
This movement completely terminalizes all 
remaining chiasmata. The dyad nature of 
each univalent is readily seen in the figure. 
In telophase I the two daughter nuclei are 
formed, and interphase I (Figure 2-2G) 



follows. The length of interphase I varies 
in different organisms. 

Each daughter nucleus undergoes the sec- 
ond meiotic division, which proceeds as ex- 
pected from mitosis. In prophase II, each 
univalent (equivalent to a chromosome with 
its two visible chromatids) contracts; at 
metaphase II (Figure 2-2H) each lines up 
at the equator of the spindle independently; 
at anaphase II the members of a dyad sepa- 
rate and go to opposite poles as monads 
(each equivalent to a single chromosome, 
since now each contains two visible chro- 
matids). Because two nuclei undergo this 
second division, four nuclei are formed at 
telophase II (Figure 2-21). Photographs 
of the meiotic process in corn can be seen 
in Figure 2-4 (pp. 20-21 ). 

Chromosomal Segregation 

Consider next the consequences of meiosis. 
The organism undergoing meiosis starts its 
existence as a zygote produced by fertiliza- 
tion involving the union of two haploid sets 
of chromosomes, one maternal and one 
paternal. When meiosis is completed the 
diploid, paired, chromosome number is re- 
duced to the haploid, unpaired, chromosome 
number. Since any postmeiotic nucleus nor- 
mally contains only one representative of any 
given pair of chromosomes present in a pre- 
meiotic nucleus, chromosome segregation 
has occurred. Two questions come to mind 
at this point. First, is the haploid set of 
chromosomes, or genome, in a gamete com- 
posed of replicas of all the chromosomes 
contributed by the female parent or of all 
those contributed by the male parent? 

For typical meiosis, the answer depends 
upon two events. The first of these is the 
manner in which the centromeres of the 
bivalents arrange themselves at the equator 
of the spindle at metaphase I. Relative to 
the poles of the spindle, each bivalent ar- 
ranges itself at the equator independently of 



Meiosis and Chromosomal Segregation 



19 



other bivalents, so that it is purely a matter 
of chance whether the copy of the ma- 
ternally-derived chromosome will go to one 
specified pole and the copy of the paternally- 
derived chromosome to the other, or vice 
versa. Consider the distribution of two bi- 
valents, for example. Since there are many 
cells undergoing meiosis in any sex organ, 
or gonad, at metaphase I, approximately half 
of these will have the two paternal univalents 
going to one pole and the two maternal uni- 
valents going to the other pole at anaphase 
I, and approximately half will have one ma- 
ternal and one paternal going to one pole 
and one paternal and one maternal to the 
other. As a result, the chromosomal con- 
tent of a pool of all the haploid nuclei pres- 
ent at the completion of meiosis will be 25% 
paternal + paternal; 25% maternal + ma- 
ternal; 25% paternal -f- maternal; 25% ma- 
ternal + paternal. Because the centromeres 
of each bivalent line up at metaphase I in 
one direction with a frequency equal to that 
in the other and because each bivalent does 
so independently of all other bivalents, we 
see that the segregation which follows occurs 
independently for different pairs of chromo- 
somes. 2 Note also, from the fate of two bi- 
valents, that 50% of haploid products have 
the same combinations of nonhomologous 
chromosomes as entered the individual in 
the parental gametes, therefore retaining the 
old or parental combinations, whereas 50% 
of haploid products carry new, nonparental 
combinations or recombinations . 

Let us defer considering the genetic im- 
plications of these conclusions until we have 
considered the second question, which also 
bears upon the maternal-paternal chromo- 
some content of gametes: our answer may 
modify the conclusions just reached. Is a 
chromosome in a gamete, in fact, a com- 
pletely uniparental replica, or has it a bi- 

2 As shown by E. E. Carothers (1921). 




B 




figure 2—3. Lily diplonema showing chro- 
matids (1-4) with different synaptic partners 
on different sides of a chiasma. (Courtesy of 
R. E. Cleland.) 



parental derivation? The latter situation 
would obtain if one segment of a gametic 
chromosome were a copy of a portion of 
one homolog and another segment a copy 
of a portion of the other homolog. 

Considerable evidence exists that some 
time between the onset of meiosis and diplo- 
nema a cytologically undetected event occurs 
which results in two of the four chromatids 
in a tetrad having segments which are bi- 
parental copies, exactly reciprocal in con- 
tent. Thus, if one biparental segment of a 
chromatid has a linear sequence that is ma- 
ternal-paternal, the other is paternal-mater- 
nal in composition. The other two chro- 



20 



CHAPTER 2 




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-a 




< 

LU 

z 

>- 

X 
u 

< 

a. 





<7 




* 
Ok 


■ 



Meiosis and Chromosomal Segregation 



21 









Q. 
O 



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^ <o O 

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!<■ ^> C 

a J <3 

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3 <s •§ 

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-5- ^ = 

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0) .2 

5 .2 
= £ 



-a ft. 



i 

n 
U 

a 






2 « ~ 



22 



( HAPTER 2 



m 
m 
P 

P 



T 



m 

rr 
P 
P 



^ 



m 



m 



HGURE 2-5. Chiasma showing paternal (p) and maternal (m) composition of strands. 
Compare with Figure 2-3B. 



matids. then, arc uniparental replicas for this 
region, one all-maternal and the other all- 
paternal. The recombinational event that 
produces the two biparental segments can 
be called an exchange, although this should 
not be taken to mean that what occurred 
was a cross-union following breakage of a 
paternal and a maternal chromatid at exactly 
corresponding positions. It is likely that 
synapsis at any level is strongest between 
strands having the same parental derivation. 
Accordingly, during diplonema of prophase 
I, when the chromatids separate in pairs, 
they should switch pairing partners at the 
point where the two biparental chromatids 
change their parental derivation. This 
switching would produce the chiasma con- 
figuration. Note in Figure 2-3B, where 
chromatid material of one parental deriva- 
tion is shown filled in whereas that of the 
other is not, that it is necessary for chroma- 
tids to change pairing partners in order to 
maintain in synapsis all the corresponding 
paternally-derived (and also all the corre- 
sponding maternally-derived) portions of the 
strands. Some of the exchanges may fail to 
produce a visible chiasma later; this could 
happen if the pairing were uniparental on one 
side and biparental on the other side of a 
point of exchange. On the other hand, every 
chiasma can be taken to represent cytological 
evidence of a prior intratetrad exchange. It 
should be noted, however, that because of 



chiasma terminalization. the position of a 
chiasma may be distal to the point where 
the two biparental chromatids changed their 
parental type. We will henceforth assume 
that chiasma terminalization is absent dur- 
ing diplonema, thereby making it possible 
to equate the position of a chiasma with the 
point of exchange. Accordingly, a tetrad 
containing one chiasma would have the pa- 
ternal (p) and maternal (m) linear consti- 
tution shown in Figure 2-5, where the cen- 
tromere is represented by C. The figure 
shows that, after one exchange, one chro- 
matid remains entirely maternal and one en- 
tirely paternal, but the other two are bi- 
parental in origin. Note again that only 
two of the four chromatids are involved in 
a single exchange. However, a tetrad nor- 
mally contains several chiasmata; this means 
that earlier each of the four chromatids had 
probably exchanged at one place or another 
with a chromatid derived from the other 
parent and consequently has a biparental 
composition. 

We can now return to the questions re- 
garding the maternal-paternal chromosome 
content of the haploid products of meiosis. 
The centromeres in a bivalent separate at 
anaphase I ; thus segregation of maternal from 
paternal centromeres occurs at the first meio- 
tic division. Then, as revealed by the loca- 
tion and number of chiasmata at diplonema, 
the monads having the maternal centromeres 



Meiosis and Chromosomal Segregation 



23 



will contain different paternal segments along 
their lengths, and the monads with the pa- 
ternal centromeres will have the comple- 
mentary maternal sections along their lengths. 
Accordingly, segregation of maternally-de- 
rived chromatid material from paternally- 
derived chromatid material occurs for cen- 
tromeric and some other regions of the chro- 
matids at anaphase I and is accomplished 
for the remaining regions of the chromatids 
at anaphase II. Since bivalents line up at 
metaphase I independently, meiosis will nor- 
mally result in the segregation of homologous 
chromosomal regions and independent segre- 
gation of chromosomal segments located on 
nonhomologous chromosomes. 

Life Cycles of Multicellular Organisms 

Since the diploid number of chromosomes is 
maintained generation after generation in 
sexually reproducing forms, it is not sur- 
prising that meiosis always occurs at some 
time in the life cycle of each sexually repro- 
ducing individual. In most animals meiosis 
comprises the last two nuclear divisions be- 
fore the mature sperm or egg is produced. 
Meiosis occurs at different times in the life 
history of different plants, but rarely just 
before the formation of gametes. Different 
species show minor variations in the details 
with which meiosis is carried out. Let us 
consider the life cycles of certain multi- 
cellular organisms which have proven to be 
especially interesting both for cytological and 
genetic investigations — that is, in cytoge- 
netics. 

1 . Drosophila melanogaster 3 

The adult stage of D. melanogaster, com- 
monly called the fruit, banana, or vinegar 
fly, is shown in Figure 2-6. Although size 
depends upon nutritional and other environ- 

; See the Appendix to this chapter, on p. 29. for 
references dealing with this and other species of 
Drosophila. 



^^5 



I I I I 1 I I II | 



-lnrr: - 





& ; . 



i>-<vt"^- 



lip 




figure 2-6. Normal (wild-type) Drosophila 
melanogaster male (A) and female (B). 
(Drawn by E. M. Wallace.) 



mental factors, the adult is usually 2-3 mm. 
long, and females are larger than males 
raised under comparable conditions. As 
found in nature, the wild-type fly has a grey 
body color and dull-red compound eyes. 
Males are readily distinguished from females 
by having sex combs on the anterior pair of 
legs and an abdomen which terminates dor- 
sally in a single broad black band (instead 
of a series of bands) and which ends ven- 
trally in a penis and claspers (instead of an 
ovipositor). 



24 



CHAPTER 2 




figure 2-7. Egg, mature larva, and pupa of Drosophila 
melanogaster. Each unit of scale equals 1 mm. (Cour- 
tesy of E. R. Balboni. ) 



The adult male is diploid and has a pair 
of testes in which spermatogonia are pro- 
duced by mitosis. When one of these sper- 
matogonia! cells enters meiosis it is called 
a primary spermatocyte. The first meiotic 
division produces two secondary spermato- 
cytes; the second meiotic division, four hap- 
loid spermatids. Each spermatid differen- 
tiates without further division into a sper- 
matozoan, or sperm cell. Note that for each 
diploid primary spermatocyte entering meio- 
sis. four functional haploid sperm are pro- 
duced at the completion of spermatogenesis. 
This is also true of males of many higher 
animals, including the frog, mouse, and man. 
Sperm are stored in the Drosophila male 
until they are ejaculated into the vagina of 
the female, where they proceed to swim into 
the female's sperm storage organs (a pair of 
spermathecae and a coiled ventral recep- 
tacle). 

The adult female has a pair of ovaries 
each of which is composed of a series of 
egg tubes, or ovarioles. At one end of the 
ovariole are diploid oogonia. By four syn- 
chronous mitotic divisions each oogonium 
produces a nest of 16 cells, one of which 
enters meiosis as a primary oocyte while 
those remaining serve as nurse cells for the 
maturing oocyte. As the oocyte grows it 
passes down the ovariole, into the oviduct 
and then the uterus. At the time it reaches 



the uterus, the egg (Figure 2-7) is usually 
no further advanced than metaphase of the 
first meiotic division. The sperm held in 
the female are released to fertilize the egg 
in the uterus, after which the first meiotic 
division continues. The two secondary 
oocyte nuclei produce four haploid nuclei, 
three of which are polar nuclei and degen- 
erate, the remaining one serving as the hap- 
loid egg nucleus. Note here, as in females 
of the frog, mouse, and man, that a single 
primary oocyte produces only one mature 
haploid egg at the completion of oogenesis. 
(In man, also, meiosis is completed after 
fertilization, the egg remaining at metaphase 
II until sperm entry.) Since the female 
Drosophila stores hundreds of sperm and 
uses them sparingly (only one sperm usu- 
ally enters the egg), a single mating can 
yield hundreds of progeny. 

At about 25° C embryonic development 
proceeds for about a day, when the larva 
hatches from the egg. After four more days 
and two moltings, the mature larva becomes 
a pupa (Figure 2-7) . After about four days 
the young adult, or imago, ecloses (hatches) 
from the pupa case. Thus, the Drosophila 
undergoes a complete metamorphosis during 
its life cycle. Although mating usually oc- 
curs during the first 24 hours of adult life, 
oviposition usually starts during the second 
day, so that the generation time is about 10 



Meiosis and Chromosomal Segregation 



25 



Pericarp 

Aleurone 

Endosperm 
Embryo 




figure 2-8. Life cycle of corn, Zea mays. See text for description. 



days. Adults can live as long as 10 weeks, 
during which a female can lay several thou- 
sand eggs. 

2. Zea mays 4 

Corn or maize, like the bean and the 
garden pea, usually produces both male and 
female sex organs on the same plant and is 
therefore monoecious ("one house"). Since 



4 See the Appendix to this chapter, on p. 29, for 
references to corn experiments and literature. 



the diploid corn plant produces male and 
female spores, microspores and megaspores 
respectively, it represents the sporophyte 
stage of the life cycle (Figure 2-8). The 
microspores are produced in tassels at the 
end of the stem. Here diploid microspore 
mother cells, or microsporocytes, each un- 
dergo meiosis to produce four haploid micro- 
spores (A), each of which develops into a 
pollen grain (B). 

Since the haploid microspores give rise to 
the haploid gametes, these two points in the 



26 



CHAP! I R 2 



life cycle comprise the beginning and end 
of the male gametophyte stage. The haploid 
microspore nucleus divides mitoticall) once. 
to produce two haploid nuclei. One of these 
nuclei elites not divide again and becomes the 
pollen tube or vegetative nucleus. The 
other nucleus divides mitotically once, so 
that the gametophyte contains three hap- 
loid nuclei (C); the two last formed func- 
tion as sperm nuclei ( D. k ) . 

Near the base of the upper branches of 
the corn plant are clusters of pistils, each 
containing one diploid megaspore mother 
cell, or megasporocyte. (The styles of the 
pistils later become the silks. ) The mega- 
sporocyte undergoes meiosis to produce four 
haploid nuclei (Eh three of which degen- 
erate (F). The remaining megaspore nu- 
cleus divides mitotically (G), as do its 
daughter and granddaughter nuclei (H). so 
that eight haploid nuclei result (I). In the 
embryo sac- (J), three of the eight nuclei 
cluster at the apex and divide to form antip- 
odal nuclei, two move to the center (polar 
nuclei), and three (composed of two syn- 
ergid nuclei and one egg nucleus) move to 
the base of the embryo sac. The pollen tube 
grows down the style to the embryo sac. 
where one sperm nucleus fertilizes the egg 
nucleus (K, L), producing a diploid (2N) 
nucleus, and the other sperm nucleus fuses 
with the two polar nuclei to produce a 
triploid (3N) nucleus. With the occurrence 
of this double fertilization the sporophyte 
stage is initiated. Mitotic division of the 
diploid nucleus (L) produces the embryo, 
whiij the triploid nucleus develops into the 
endosperm. The surface cells of the endo- 
sperm comprise the aleurone, containing 
aleurone grains and oil; the remaining endo- 
sperm cells contain starch. The endosperm 
is gradually digested to nourish the em- 
bryo and seedling. The outer surface of 
the kernel is the pericarp, diploid tissue 
derived from the maternal sporophyte. In 



other words, a corn kernel has its pericarp 
produced by one Sporophyte and its remain- 
ing tissue h\ the sporophyte o\ the next gen- 
eration. Development from embryo sac to 
mature kernel requires about eight weeks 
(during which the antipodal and synergid 
nuclei or their deseendents degenerate); de- 
velopment from the kernel to the mature 
sporophyte occurs in about four months 
(during which the first leaf completes the 
digestion of the endosperm). 

3. Neurospora crassa "' 

Neurospora ("nerve spore") is a bread 
mold whose haploid vegetative stage is com- 
posed of threads or hyp hue which interweave 
to form a mass, the mycelium. The hyphae 
branch and fuse. Since the cell walls which 
partition the hypha into cells are incomplete, 
the cytoplasm of the filament is continuous. 
Each hyphal cell is multinucleate. 

Cultures can be propagated asexually by 
transplanting pieces of mycelium, or by 
spores (conidia) which contain one or sev- 
eral haploid nuclei. Sexual reproduction 
(Figure 2-9) requires the participation of 
molds of different mating type which pro- 
duce fruiting bodies. A haploid nucleus 
from one mating type enters and divides 
mitotically a number of times in the fruit- 
ing body of the opposite mating type. These 
haploid nuclei pair with haploid nuclei of 
the fruiting body and fuse to produce diploid 
zygotic nuclei. Each zygotic nucleus then 
undergoes two meiotic divisions to produce 
four haploid nuclei; each of these divides 
once mitotically to form a total of eight hap- 
loid nuclei. The cytoplasm is then parti- 
tioned to form eight ovoid haploid asco- 
spores which are contained in a thin-walled 
sac, the ascus. The mature fruiting body may 
contain 300 asci, from which the ascospores 
are released and carried in the air. Upon 

•' See Appendix on p. 29. tor references to experi- 
mental methods, results, and literature. 



Meiosis and Chromosomal Segregation 



27 



germination the haploid ascosporc divides tcrent strains of the same or different mating 
mitotically and grows to produce the my- type will fuse to form a heterocaryon, hyphal 
celium. Sometimes the hyphae of two dif- cells containing the nuclei of both strains. 



Mycelium 
(mating type A) 




Mycelium 
(mating type Bl 



scospore M 

rmination '"Y'fw 



Ascus with 
ascospores (N) 



figure 2-9. Life cycle of Neurospora. See text for description. 



SUMMARY AND CONCLUSIONS 

Meiosis involves two successive, essentially mitotic divisions modified by the occurrence 
of synapsis and of chiasmata formation during prophase I. and the nonoccurrence of 
chromosomal replication during interphase I. Any particular chromosome in a genome 
of a gamete has an equal chance of having a replica of a maternally- or a paternally- 
derived centromere (because of the random manner in which different bivalents align 
themselves on the spindle at metaphase I) and usually contains replicas of segments 
originally derived from the other parent (as revealed by chiasmata). As a conse- 
quence of meiosis the original pair of genomes becomes single in the gametes. Not 
only do homologous chromosomal segments segregate, but chromosomal segments of 
nonhomologs segregate independently. 



28 



I II MTER 2 




EDMUND B. WnLSON (1856-1939), American 
cytologist. (From Genetics, vol. 34. p. I. 
1949 



REFERENCES 

Carothers. E. E.. "Genetical Behavior of Heteromorphic Homologous Chromosomes 
of Circotettix (Orthoptera)." J. Morphology. 35:457-483. 1921. 

Darlington. C. D.. and Bradshaw. A. D. (Eds.). Teaching Genetics in School and Uni- 
versity. Edinburgh: Oliver & Boyd. 1963. 

De Robertis. E. D. P.. Nowinski. W. W.. and Saez. F. A.. General Cytology, 3rd Ed.. 
Philadelphia: Saunders. 1960. 

Jehle. H.. 'intermolecular Forces and Biological Specificity," Proc. Nat. Acad. Sci.. 
U.S.. 50:516-524. 1963. 

Lewis. K. R... and John. B.. Chromosome Marker, London: J. and A. Churchill Ltd.. 
1963. 

McLcisch. J., and Snoad, B.. Looking at Chromosomes. New York: St. Martin's Press, 
1958. 

Rhoades. M. \l.. "Meiosis." pp. 1-75. in The Cell. Vol. 3, Meiosis and Mitosis. Brachet, 
J., and Mirsky. A. E. (Eds.). New York: Academic Press, 1961. 

Sutton. W. S.. 1903. "The Chromosomes in Heredity." Biol. Bull.. 4:231-251. Re- 
printed in Great Experiments in Biology, Gabriel. M. L.. and Fogel, S. (Eds. i. 
Englewood Cliffs. N.J.: Prentice-Hall. 1955, pp. 248-254; also in Classic Papers 
in Genetics. Peters. J. A. (Ed.), Englewood Cliffs, N.J.: Prentice-Hall. 1959, pp. 

27-41. 

Swanson, C. P.. Cytology and Cytogenetics, Englewood Cliffs. N.J.: Prentice-Hall. 1957. 

Van Beneden. E.. 1883. "Researches on the Maturation o\ the Egg and Fertilization." 
translated in Great Experiments in Biology, Gabriel, M. L.. and Fogel. S. (Eds.). 
Englewood Cliffs. N.J.: Prentice-Hall. 1955. pp. 245-248. 

Wilson, E. B.. The Cell in Development and Heredity. 3rd Ed., New York: Macmillan. 

1937. 



Meiosis and Chromosomal Segregation 2?) 

APPENDIX 

Drosophila 
Bibliographies on most, if not all, investigations with all species of Drosophila through 
1962 are found in source I (which includes subject indexes), and the more recent 
work in 5. The lite cycle, culture, cytological and genetic experiments for the class- 
room are given in 3, 4, 6, 8. All aspects of Drosophila biology are treated in detail in 3. 
Mutants found until about 1942 are described in 2. Many mutants found since then 
are described in 5, or referred to in 1 . A new and complete compilation is in prepara- 
tion (7). Stock lists of various Drosophila species maintained in different laboratories, 
addresses of Drosophila workers, research and teaching notes are given international 
presentation in the more or less annual bulletin of 5. 

1. Bibliography on the Genetics of Drosophila: Part I by H. J. Muller (Edinburgh: 
Oliver and Boyd, 1939, 132 pp.). Parts II, ///, and IV by I. H. Herskowitz (Ox- 
ford: Alden Press, 1953, 212 pp.; Bloomington: Indiana University Press, 1958, 
296 pp.; and New York: McGraw-Hill, 1963, 344 pp., respectively). 

2. Bridges, C. B., The Mutants of Drosophila melanogaster (completed and edited by 
Brehme, K. S.), Washington, D.C.: Carnegie Institution of Washington, Publ. 552, 
257 pp., 1944. 

3. Demerec, M. (Ed.), The Biology of Drosophila, New York: J. Wiley & Sons. 632 
pp., 1950. Xerographed by University Microfilms, Inc.. 313 N. 1st Street, Ann 
Arbor, Michigan. 

4. Demerec, M., and Kaufmann, B. P., Drosophila Guide, 7th Ed. Washington, D.C.: 
Carnegie Institution of Washington, 47 pp., 1961. 

5. Drosophila Information Service (E. Novitski. Editor, Dept. of Biology, Universit\ 
of Oregon, Eugene, Ore.). 

6. Haskell, G., Practical Heredity with Drosophila, Edinburgh and London: Oliver 
and Boyd. 124 pp., 1961. 

7. Lindsley, D. L., and Grell, E. H., The Mutants of Drosophila, Washington, D.C.: 
Carnegie Institution of Washington, 1965. 

8. Strickberger, M. W., Experiments in Genetics with Drosophila, New York: J. Wiley 

& Sons, 144 pp., 1962. 

Zea 

1. Kiesselbach, T. A., "The Structure and Reproduction of Corn," Univ. Nebraska 
Coll. Agric, Agric. Exp. Sta. Res. Bull., No. 161, 1949. 

2. Maize News Letter (M. M. Rhoades, Ed.. Dept. of Botany, Indiana University, 
Bloomington, Ind.). 

3. Sprague, G. F., Corn and Corn Improvement, New York: Academic Press, 1955. 

4. Weijer, J., "A Catalogue of Genetic Maize Types Together with a Maize Bibliog- 
raphy," Bibliographica Genetica, 14:189-425, 1952. 

Neurospora 

1. Bachmann, B., and Strickland, W. N.. Neurospora Bibliography and Index. New 
Haven: Yale University Press, 1965. 

2. Fincham. J. R. S., and Day, P. R., Fungal Genetics, Oxford: Blackwell Scientific 
Publications. 1963. 

3. Ryan, F. J., "Selected Methods of Neurospora Genetics," Methods in Medical Re- 
search. 3:51-75, 1950. 

4. Wagner, R. P., and Mitchell, H. K... Genetics and Metabolism, 2nd Ed., New York: 
J. Wiley & Sons, 1964. 



30 CHAPTER 2 

QUESTIONS FOR DISCUSSION 

2.1. Can sexuall) reproducing organisms reproduce asexually? Is the reverse true? 
Explain. 

2.2. What arc the similarities and differences between mitosis and meiosis? 

2.3. Suppose the meiotic process had never evolved. What do you think would have 
been the consequence? 

2.4. Certain unusual chromosomes are rings rather than rods. Could a ring chromo- 
some, during meiosis. have any difficulty that a rod chromosome would not have? 
Explain. 

2.5. How main hivalents are present at metaphase I in man? Corn? The garden pea? 

2.6. Discuss the statement: During meiosis. each segment of a chromosome segregates 
independently of its homologous segment and of all other chromosome segments. 

2.7. Argue against the hypothesis that the physical hasis of genetic material lies in 
the chromosomes. 

2.8. What do you suppose happens during meiosis in individuals possessing an odd 
numher of chromosomes? 

2.9. Suppose an animal has a diploid chromosome number ol six. What proportion 
of all its gametes would receive replicas of all the centromeres originally derived 
from the father? From the mother? From either the father or mother? From 
both the father and mother? 

2.10. What do you suppose is meant by the expression "first division segregation"? 
"Second division segregation"? If exchanges giving rise to chiasmata can occur 
at a variety of positions along the chromosome, under what circumstances can a 
given chromosomal segment undergo first division segregation? Second division 
segregation? 

2.11. If you saw a single cell at metaphase. how could you tell whether the cell was 
undergoing mitosis, metaphase I, or metaphase 11 at the time it was fixed and 
stained? 

2.12. What advantages do the following organisms offer lor the study of cytology 
and/or of the genetic material: Drosophila? Corn? Neurospora? 

2.13. What are the major differences between spermatogenesis and oogenesis in 
Drosophila? 

2.14. Suppose a pair of originally identical homologous chromosomes having arms of 
equal length was changed permanently so that one gained a large knob at one end 
and the other a small knob at the opposite end. Draw the appearance of these 
new homologs (a) at mitotic prophase and (b) at diplonema. Draw the appear- 
ance of all the monads in part b as seen at telophase II. 

2.15. Is the female sex cell of man or Drosophila ever haploid? Explain. 

2.16. How does a monad at diplonema differ from the same monad at telophase II? 

2.17. Can you suggest any functions which the polar nuclei in Drosophila oogenesis 
may serve? 

2.18. Is mitosis in triploid endosperm expected to be normal? Why? 



( 'hapter 3 

SEGREGATION OF ALLELES 



Ti 



|he similarities and the differ- 
ences in phenotypes, both 
among offspring and between 
them and their parents, have led us to postu- 
late the existence of genetic material. Since 
this material is supposedly transmitted from 
generation to generation, we may be able 
to learn more about the transmissive prop- 
erties of the genetic material by studying the 
traits that recur in lines of descent. This 
area may be called "transmission genetics." 
We could investigate the genetic material 
either in lines reproducing asexually or in 
lines (like the beans already discussed) that 
reproduce sexually by self-fertilization. How- 
ever, instead of taking either of these paths 
of investigation, both of which deal with 
pure lines, let us study the transmission ge- 
netics of organisms reproducing sexually by 
cross-fertilization. In the experimental work 
described henceforth, it can be assumed, 
unless stated to the contrary, that appro- 
priate precautions have been taken to assure 
that the phenotypic similarities and differ- 
ences described are genotypic in origin and 
are not due to variations in environmental 
conditions. 

Different strains of a cross-fertilizing ani- 
mal or plant often show phenotypic differ- 
ences with respect to a given trait. For 
example, with respect to height, one line 
might be short and the other, tall; or with 
respect to color, one line might be red and 
the other, white. The question to be raised 
now is what will happen phenotypically in 
the offspring when two lines showing dif- 
31 



ferent alternatives for the same trait are 
crossed? Will such results tell us anything 
about the genetic material? 

Consider some specific experiments that 
can be performed with the garden pea, 1 first 
with respect to what should be done and 
why it should be done. Then we can exam- 
ine the results obtained and discuss what 
they reveal regarding the genetic material. 

The garden pea is well suited for this work 
because it is simple and inexpensive to raise 
and has a generation length short enough to 
permit the study of a number of successive 
generations. Although garden peas are nor- 
mally self-fertilizing, they can also be cross- 
fertilized; in fact, the experimenter can con- 
trol all mating by simple and appropriate 
techniques. Moreover, there are numerous 
strains which differ phenotypically with re- 
gard to different traits. It is first necessary, 
of course, to self-fertilize each strain for 
several generations and check the pheno- 
types, to be sure that pure lines are being 
used. 

Which pure lines should one cross to- 
gether? Since we do not know what to ex- 
pect in the offspring, we should avoid using 
lines whose traits, for environmental reasons, 
are so variable that a phenotype in one line 
also occurs in the other (which was the sit- 
uation in the bean strains studied in Chapter 
1 ). Such phenotypic overlaps could prevent 
us from deciding from the phenotypes what 
genotypes were present. Consequently, we 
should select for study only those strains 
showing a sharp, nonoverlapping, easily de- 
tected difference. For simplicity, we should 
use only strains having a single major dif- 
ference. We should study only lines that 
can be successfully cross-fertilized in both 
directions; that is, the male gamete should 
be furnished sometimes by the one line and 
other times by the other line. Such re- 

1 Based upon the experiments of G. Mendel (see 
p. s-9). 



32 



(MAI' I I K 3 



ciprocal matings arc desirable in order to 
determine whether it makes any phenotypic 
difference to the progenj upon which paren- 
tal line the offspring start developing (note 

that pea seeds form on the maternal parent ). 

All crosses should also be fully fertile; 
that is. the parental lines should be hardy 
plants that grow vigorously and produce full 
sets of seed capable of growing to maturity, 
not only when self-fertilized but when 
crossed to each other reciprocally. If this 
precaution is not taken insufficient numbers 
of offspring may be obtained or, more im- 
portant, the offspring observed may com- 
prise a biased sample of those starting de- 
velopment. Deaths that occur between the 
time of fertilization and the time the pheno- 
typic observations regarding the offspring 
are made may not occur at random. Differ- 
ential viability for different genotypes could 
cause one to miss, or underestimate the fre- 
quency of, certain phenotypes; this would 
give misleading results with regard to geno- 
types, especially on our present view that 
the genetic material is transmitted at the 
time the new organism starts its existence, 
i.e., at the time of fertilization. Of course, 
accurate records of lineage and of parental 
and offspring phenotypes must be kept. 

Two strains of garden pea, one producing 
colored flowers and the other colorless 
flowers, satisfy the prerequisites discussed. 
Cross-fertilizations are made reciprocally 
between pure-line colored flowers and pure- 
line colorless flowers, these individuals serv- 
ing as the parents of the first generation 
(P,). The offspring seeds are planted and 
the color of the flowers produced is scored. 
All the offspring, which comprise what may 
be called the first filial generation (Fi), are 
phenotypically uniform, having colored flow- 
ers just like one of the P,. The F, results 
are the same for the reciprocal matings. In 
the discussion that follows in this and subse- 
quent chapters it will be correct to assume 
that all crosses were made reciprocally and 



produced identical results, unless a statement 
to the contrary is made. 

What can one conclude about the genetic 
material from these results? Let us use 
symbols as a shorthand method of repre- 
senting the genetic material — C for the ge- 
netic material whose action produces colored 
flowers, present in all members of the col- 
ored-flowered pure line, and r for the genetic 
material producing colorless flowers, present 
in all the colorless-flowered pure-line indi- 
viduals. All F 1 individuals must contain C 
since they produce colored flowers. What 
has happened to c? Has it failed to be trans- 
mitted? 

More may be learned by permitting the 
Fi plants with colored flowers to serve as 
P L > (parents of the second generation) and 
reproduce by self-fertilization to yield F 2 
progeny. When this is done, and sufficient 
numbers of F 2 are obtained from each P 2 
plant, one finds among the offspring of every 
P 2 that some are colored and some are white. 
In terms of genetic material, these F L > must 
carry, respectively, C or c. It is no surprise 
that some F L . contain C, but where did the c 
come from? In these cases, one could at 
first suppose either that c arose spontaneously 
from some non-genetic origin or that C 
mutated to c. The first explanation can be 
bypassed in view of the previous assump- 
tions (p. 10) that genetic material can arise 
only from pre-existing genetic material, and 
that this material is self-reproducing (self- 
replicating). The second explanation can 
be eliminated by the observation for the 
pure line containing C that mutations to c 
are thousands of times rarer than the occur- 
rence of c among the F L >. If the P-(F X ) are 
genotypically like pure-line C individuals, as 
assumed, mutation cannot be the explana- 
tion for the difference in breeding behavior 
between P,C and P L C. 

The results of the bean experiments in 
Chapter 1 are consistent with the view that 
the genetic material in any individual is a 



Segregation of Alleles 



33 



single indivisible unit. In the absence of a 
simpler explanation for the present findings 
with peas, it seems necessary to postulate 
that the genetic material is not always com- 
posed of a single indivisible unit. The ap- 
pearance of c in F 2 can be explained by 
making the more complex assumption that 
each P L .(Fi) individual contains not only C 
but c as well; in other words, that in some 
individuals the genetic material contains two 
units. Let us use the word gene to refer to 
a unit or restricted portion of the genetic 
material. But, if it is assumed that there 
is a pair of genes in each P L ., then all other 
individuals in our experiment must be as- 
sumed to have a pair of genes, too. For, 
in science, we adhere to the law of parsimony 
{Occam's rule), which states that one must 
use the minimal number of hypotheses or 
assumptions to explain a given set of obser- 
vations. Instead of having some individuals 
with paired genes and others without pairs, 
then, all are assumed to have a pair of genes 
in their genetic material. Accordingly, the 
two pure lines and the Pi must have been 
CC and cc, and all Ft must have been Cc. 
Those F 2 which are colorless must be cc. 
Attention is called to the individuals in F 2 
that are cc. These have colorless flowers 
phenotypically identical with those of the 
original colorless pure line used in the P^ 
In fact, crosses of F 2 colorless individuals 
either with themselves or with any other 
colorless individual (F 2 , or pure line) pro- 
duce all colorless progeny. In other words, 
F- cc individuals are genotypically just as 
pure with respect to the trait under consid- 
eration as are pure-line individuals. This 
is true despite the fact that both c's in the 
F 2 had been carried in F] individuals in 
which C was the other member of the pair 
of genes. We conclude, therefore, that when 
c is transmitted to the F 2 , it is uncontam- 
inated, or untainted, by having been in the 
presence of C in the F t , even though c had 
not been expressed in any noticeable way in 



the phenotype of the F, individuals. We 
can generalize this conclusion and state that 
the nature and transmission of any gene is 
uninfluenced by whatever its partner gene 
may be. The members of a gene pair are 
said to be alleles (partner genes), a term 
also applied to alternative forms of a given 
gene. 

Since each P 2 produced colored and color- 
less F 2 offspring, each P 2 had the genotype 
Cc composed necessarily of C from the 
CC P, and c from the cc P^ This specifies 
that one and only one member of a pair of 
genes in a parent is transmitted to each 
offspring, so that in the transmission process 
the members of a parental pair of genes must 
become separated, or segregated, from each 
other. The paired, or diploid, status of the 
genes becomes unpaired, or haploid, during 
transmission; but diploidy is restored in the 
offspring because a haploid genotype is con- 
tributed to it by each parent. 

Accepting the hypothesis that paired genes 
are segregated by the time they are trans- 
mitted to progeny, are the two alleles in a 
parent equally likely to be transmitted to 
offspring? The F 2 produced by self-fertiliza- 
tion of Fx Cc demonstrate that both genes 
of a given individual are transmissible. Let 
us test the hypothesis that both members of 
this pair of alleles are equally transmissible. 
If so, the Fx male parent (or part) would 
contribute C one half the time and c the 
other half; similarly the Fi female parent 
(or part) would contribute C half the time 
and c the other half. Finally, assume di- 
ploidy is restored at random; that is, the 
haploid gene contributed to the offspring by 
one parent is uninfluenced by the haploid 
gene contributed by the other parent. Ac- 
cordingly, an offspring that receives C from 
the male (50% of offspring) will have an 
equal chance of receiving C or c from the 
female, so that of all offspring 25% will be 
CC and 25%- Cc. Those offspring receiv- 
ing c from the male (50% of offspring) will 



34 



( II \l'l p.r 3 



P CC x cc (Cross - fertilization) 

I J 

O. all C all c (Gametes) 



all Cc 



Cc x Cc (Self-fertilization of F, 

A A 

Vi C, Vt c Vi C, Vic 



Male gametes 
Vi C 7* c 



»/« cc 


'/« Cc 


'/« Cc 


'/« cc 



Female ' 7 C 

gametes ,/ 
3 Vi c 



or ' 4 CC »/j Cc V, cc 

when * t 

P 3 self- breeds breeds breeds 

fertilized like like like 

P, CC P Cc P cc 



figure 3—1. Genotypic model proposed to 
explain the phenotypic results of certain crosses 
involving colored and colorless flowers in pea 
plants. 



also have an equal chance of receiving C or 
c from the female, so that the contribution 
to all the offspring genotypes will be 25% 
Cc and 25% cc from this source. On this 
basis the F L . would be predicted to contain 
25% of individuals that arc CC. 50% that 
are Cc, and 25% cc. This expectation can 
be expressed as relative frequencies in 
several ways: % CC: '{> Cc % cc, or 1 CC: 
2 Cc : 1 cc, or .25 CC : .50 Cc : .25 cc. As al- 
ready reasoned CC and Cc are phenotyp- 
ically indistinguishable, having colored flow- 
ers, so that phenotypically 75% of the F 2 
would be colored and 25% would be color- 
less. What is their relative frequency in the 
F 2 actually observed? 



Although a penny has in theory a 5() r i 
chance of falling head up and a 50% chance 
oi falling tail up, obviously a sufficiently 
large Dumber of tosses is required to obtain 
approximately 5095 heads, 509? tails. Sim- 
ilarly, an accurate test of the theoretical ex- 
pectation of 75% colored and 25% color- 
less will be obtained only if a sufficiently 
large sample of offspring is scored. Instead 
of scoring the offspring of just one P 2 , then, 
the results for the offspring of all P 2 should 
be totalled. When this is done, the actual 
F 2 results (among 929 plants. 75.9% are 
colored and 24. 1 % colorless ) are very close 
to the expectation. 

It should be emphasized that the concept 
of paired, untaintable, segregating genes has 
not depended upon obtaining any particular 
phenotypic ratio for the F 2 . Granting these 
characteristics of the genetic material, ob- 
taining or not obtaining the phenotypic ratio 
:; , colored to % colorless merely tests the 
suppositions ( 1 ) that there is an equal 
chance for offspring to receive either haploid 
product of gene segregation from a parent, 
and (2) that the haploid products from dif- 
ferent parents come together at random to 
restore the diploid condition. 

If all the assumptions so far made about 
genetic material are correct, the 75% of F 2 
that are colored should have one of two 
genotypes: Vs of them should be CC, breed- 
ing like pure-line CC individuals, and % 
should be Cc. breeding like the F, Cc indi- 
viduals. Accordingly, each F 2 colored plant 
is permitted to self-fertilize and, in fact, very 
nearly K produce only colored F :i . whereas 
-;. produce F ;! of both colored and colorless 
types. The theoretical genotypic ratio ex- 
pected in the F L ., % CC:' 1 ^ ^ c l A cc > is, in 
this way, fully confirmed in experience. The 
gene model proposed to explain these pheno- 
typic results is summarized in Figure 3—1. 
It is convenient to introduce two additional 
terms at this time. A homozygote is an 
individual that is pure with respect to the 



Segregation of Alleles 



35 



paired genes in question, like CC or cc, 
whereas a heterozygote, or hybrid, is im- 
pure in this respect, like Cc. 

An independent test of all the genetic 
hypotheses presented so far can be made in 
the following way. F, colored plants are 
crossed with colorless plants, this cross being 
symbolized genetically: F, Cc X cc. As the 
result of segregation half of the offspring 
should receive C and half c from the Cc 
parent, and all should receive c from the cc 
parent. The genotypes of the offspring from 
this cross should be, theoretically, Cc 50% 
of the time and cc 50% of the time, and the 
expected phenotypic ratio should be, there- 
fore, 1 {» colored: l / 2 colorless. This expecta- 
tion is fulfilled experimentally (85 colored: 
81 colorless). 

Are the principles just established gen- 
erally applicable? Thus far they apply 
strictly only to the genetic determination of 
flower color in garden peas. All these ideas 
can be tested six additional times, using six 
other traits in garden peas, each of which 
occurs in two clearcut alternatives and ful- 
fills the prerequisites for suitability already 
described. In each case, when two appro- 
priate pure lines are crossed, the Fi hybrids 
produced are phenotypically uniform, as be- 
fore. Moreover, self-fertilization of the F, 
produces Fj in approximately the expected 
1:2:1 genotypic ratio. 

Recall that the Cc phenotype is indistin- 
guishable from CC. In Cc individuals the 
phenotypic expression of c is masked by the 
expression of C. The ability of a gene to 
express itself phenotypically in the presence 
of a different allele is described in terms of 
dominance. In the case of flower color, C 
produces a dominant effect when present 
with c, whose effect is, accordingly, recessive. 
It should be emphasized that our concept of 
the gene does not depend upon the occur- 
rence or nonoccurrence of dominance. In- 
deed, testing our genetic postulates has been 
made more complicated by the fact that the 



effect of C is, for all intents and purposes, 
completely dominant to that of c. The Fi Cc 
expressed only C, and the presence of c was 
detected only by breeding F, individuals, and 
observing cc progeny. Only by breeding the 
colored F L > were we able to determine that ' :{ 
were CC and % Cc. Dominance, then, re- 
fers to the phenotypic expression of genes 
in heterozygous condition and has no rela- 
tion to their integrity, replication, or mech- 
anism of transmission. For convenience, 
dominant and recessive will be used here- 
after to refer to genes, but the precise mean- 
ing of these terms should always be kept 
in mind. 

As mentioned, six other traits have been 
used to test the general applicability of the 
gene concept. In each case it happened that 
one allele was dominant to the alternative 
one in the hybrid. It is tempting to con- 
clude that dominance is a universal phe- 
nomenon since it was found to hold for 
each of seven different traits based on genes 
in the garden pea. Before making this de- 
cision, however, examine the results with re- 
gard to feather color of breeding certain 
chickens. Here black X white produces 
blue-gray F]. Mating two blue-gray Fx pro- 
duces in F_. ] 4 black, 1 /> blue-gray, and % 
white. In this case complete dominance 
does not occur, so that complete dominance 
is not a rule for the phenotypic expression 
of alleles in heterozygotes. Whenever domi- 
nance is incomplete or absent, genotypes can 
be stated with certainty from a knowledge 
of phenotypes. 

Cross-fertilization made it possible to 
show that genes occur as pairs, which be- 
come unpaired after segregation, then re- 
combine to form pairs in the offspring. In 
other words, the view that the genetic ma- 
terial contains separable paired units is based 
upon the recombination which these units 
undergo in cross-fertilizing species. The 
meaning of the term genetic recombination 
ought to be considered at this point. The 



36 



CHAPTER 3 



genetic units themselves are not required to 
undergo novel changes (mutations) when 
undergoing recombination. That is. the 
types of genes present in a genetically re- 
combinant individual existed before recom- 
bination. Given an individual whose gene 
pair is AA', segregation followed by self- 
fertilization may produce A A' again. This 
genotype is not considered to be a genetic 
recombination, but rather a reconstitution 
of the original arrangement of the units. 
The self-fertilization under discussion may 
also produce A A or A' A'. These represent 
two new genetic combinations relative to the 
parental combination, and are considered to 
be genetic recombinations. Accordingly, 
when events lead to the production of "old" 
combinations and "new" combinations of 
genes, only the latter type of grouping is 
called genetic recombination. This usage is 
reasonable in view of the importance that 
new combinations have for our understand- 
ing of genetic material (it was possible to 
derive the principle of gene segregation only 
because new combinations of genes were 
produced via sexual reproduction). Ac- 



O NORMAL FEMALE V 

I I NORMAL MALE r>f 

\/ UNKNOWN 

9 AFFECTED W 

I AFFECTED ^ 
f") — i— I Marriage Line 



QrO 



sn 





Offspring Line 

Offspring, in order 
of birth | I, to r. ) 

Dizygotic Twins 
Monozygotic Twins 



FIGURE 3-2. Symbols used in human pedigrees. 



cordingly, genetic recombination should be 
identified with any reassortment or regroup- 
ing of genes which results in a new arrange- 
ment of them. Any process, like segregation 
or fertilization, that has the potential of pro- 
ducing new arrangements of genetic units is. 
therefore, a mechanism for genetic recom- 
bination. 

The phenotypic results of the experiments 
discussed in Chapter 1 led us to hypothesize 
the existence of genetic material which is 
self-replicating, mutable, and transmissible. 
The pea plant experiments reveal that the 
genetic material can be divided into a pair 
of units by means of the operation or tech- 
nique of genetic recombination. Techniques 
or operations other than recombination can 
be employed to study the nature of the ge- 
netic material. Should other operations also 
divide the total genetic material into units, 
this would not necessarily mean that the 
units revealed by different operations are 
equivalent. Thus, to use a nongenetic anal- 
ogy, a book (equivalent to the total genetic 
material ) can be described operationally in 
terms of words, letters, numbers, pages, il- 
lustrations, and so forth. Each operation 
reveals something about the book, but the 
different units by which it is described are 
necessarily neither identical nor mutually 
inclusive. 

What bearing has the discovery of segre- 
gating alleles upon the hypothesis that the 
chromosomes represent genetic material? 
Both genes and genomes are unpaired in 
gametes and paired in zygotes. Can a 
genome be the physical basis of a gene? 
Though the gametes contain a single genome, 
this is usually constituted (ignoring the ex- 
changes leading to chiasmata, for the mo- 
ment) of replicas of some maternal and 
some paternal chromosomes. Since segre- 
gated genes are uncontaminated by their 
former alleles, being just as pure in the 
gametes as they were when they entered the 
organism at fertilization, one can reject the 



Segregation of Alleles 



37 



CM^^ 



O 



o 



^t^6 ] u66t666 t o6 



i6W55c57) c56i 65 



figure 3-3. A pedigree of albinism in man. 



65 



hypothesis that a genome is the material 
equivalent of a gene. Nor can an entire 
chromosome be identified with a gene, since 
any given gametic chromosome is typically 
constituted of some maternally-derived and 
some paternally-derived parts, as a conse- 
quence of the exchanges leading to chias- 
mata. However, the possibility still remains 
that the gene is associated with a particular 
segment of a chromosome which is so short 
that it cannot undergo an exchange leading 
to a chiasma with a corresponding segment 
on a homologous chromosome. Such a seg- 
ment would always retain its pure maternal 
or pure paternal constitution after meiosis. 
Accordingly, the maximal size of such an 
uncontaminable segment would equal the 
maximal size of the gene. 

* Segregation of Alleles in Man 

The genetic principles discovered in garden 
peas should also hold equally well for any 
other sexually reproducing species, including 
man. Because we naturally have great in- 
terest in ourselves, let us consider some 
human traits which may be based upon the 
action of a single pair of genes. The study 
of human genetics is complicated by the fact 
that, unlike other species of plants and ani- 

* Throughout this book, the asterisk indicates an 
optional chapter or section. 



mals. our species is not bred experimentally. 
Because of this scientific difficulty special 
methods of investigation have to be em- 
ployed. These include the pedigree, family, 
population, and twin methods. The present 
discussion deals primarily with the first two 
of these methods. 

The pedigree method uses phenotypic 
records of families (family trees or gene- 
alogies) extending over several generations. 
In recording pedigrees certain symbols are 
used by convention (Figure 3-2). In a 
pedigree chart a square or { represents a 
male, a circle or 9 represents a female: 
filled-in symbols represent persons affected 
by the anomaly under discussion. In con- 
trast, the family method utilizes the pheno- 
types only of parents and their offspring: 
that is. it uses data that span only one gen- 
eration. 

Albinism, or lack of melanin pigment, is 
a rare disease which occurs approximately 
once per 20.000 births. Studies of families, 
and of pedigrees like the one in Figure 3-3. 
yield the following facts, each of which is 
discussed relative to the hypothesis that 
albinism occurs in homozygotes for a reces- 
sive gene. a. Alternative hypotheses that 
albinism is due to a completely or a partiallv 
dominant gene soon prove untenable: 

1. Both of the parents of albinos mav be 



18 



( MAI'I'ER 3 



nonalbino, i.e., normally pigmented. This 
may be explained bv the occurrence of .1 
homozygous albino child (aa) from the 
marriage of two nonalbino heterozygotes 
(Aa ■ Aa). 

2. The trait appears most frequentlv in 
progenj sharing a common ancestor. In 
Sweden and Japan, although the percentage 
of all marriages between cousins is less than 
595 in the general population, it is 20 to 
509? among the parents of albino children. 
Since albinos are rare, so is the a gene. Ac- 
cordingly, the chance of obtaining homozy- 
gous, aa. individuals is relatively slight if 
the parents are unrelated, because even when 
the first parent is A a or aa, the second parent 
is most likely to be A A. On the other hand. 
if once again the first parent is A a or aa, 
marriage with a related individual makes it 
much more likely that the second parent 
carries an a received from the ancestor held 
in common with the first parent. 

3. What relative frequencies of nonalbino 
and albino children are expected when one 
tallies all the children in two-child families 
in which albinism appears in the progeny 
even though both parents are nonalbino? 
Based on the hypothesis under consideration, 
the parents must be Aa X A a. From such 
a marriage, the chance that a given child is 
nonalbino is :: 4 and that it is albino 1 4- 
Each child produced from such a marriage 
has these same chances for nonalbinism and 
albinism, chances which are independent of 
the genotypes (or phenotypes) of the chil- 
dren preceding or following it in the family. 
Accordingly, of all two-child families whose 
parents are Aa, % have the first child non- 
albino. and of these % also have the second 
child nonalbino. Thus, % 6 of all two-child 
families from heterozygous parents are ex- 
cluded from our sample, since both children 
are normally pigmented. Our sample in- 
cludes the following, however: families 
whose first child is normal (% ) and second 
child is albino ( ', ). making up % 6 ( % of 



;; , ) o{ all tWO-child families; families where 
the reverse is true ( :; , o\' ',). comprising 
another ; ,,, oi all two-child families; and 
families in which both children are albino 
( ' 1 of ' , ). which make up ' , o\ all two- 
child families. On the average, then, everv 
seven albino-containing families scored 
should contain six normal children (three 
from each of the two kinds of families con- 
taining one albino) and eight albinos (three 
from each of the two kinds of families con- 
taining one albino, and two from each family 
containing two albinos), so that the ratio 
expected is 3:4 as nonalbino: albino. The 
ratio actually observed closely approximates 
the expected one. 

The observed proportions of nonalbino 
and albino children in families of three, or 
of four, or of more children from normal 
parents also fit the expected proportions cal- 
culated in a similar manner. 

4. Marriages between two albinos pro- 
duce only albino children, as expected ge- 
netically from aa X aa. 

5. Twins arising from the same zygote 
(monozygotic or identical twins) are either 
both albino or nonalbino. Since ordinarily 
such twins are genetically identical, both arc 
expected to be normal, AA or Aa, or albino. 
aa. Twins arising from different zygotes 
(dizygotic, nonidentical, or fraternal twins) 
are no more likely to be the same with re- 
spect to albinism than any two children of 
the same parents. 

These evidences offer clear proof that an 
albino person is usually homozygous reces- 
sive for a single pair of segregating genes. 

The anomaly of woolly hair is a rare trait 
in Norwegians. After studying pedigrees, 
woolly hair can be attributed to the presence 
of a rare dominant gene, represented by W. 
When woolly-haired individuals (Ww) 
marry normal-haired individuals (tin), it 
is expected and found that approximate!} 
50% of children have woolly hair and 50% 
have normal hair. Note that the affected 



Segregation of Alleles 



39 



parent is represented as a heterozygote. Be- 
cause the trait is so rare, and because, bar- 
ring mutation, both parents of a homozygous 
WW child would have to have woolly hair, 
WW probably does not occur. 

Finally, consider the genetic basis for cer- 
tain kinds of anemia. Two special kinds 
occur among native or emigrated Italians. 
One type, severe and usually fatal in child- 
hood, is called thalassemia major or Cooley's 
anemia; the other type, a more moderate 
anemia, is called thalassemia minor or micro- 
cytemia. Pedigree and family studies show 
that both parents of t. major children have t. 
minor, and all the data support the hypoth- 
esis that individuals with t. major are homo- 
zygotes (tt) for a pair of genes, and that 
persons with t. minor are heterozygotes (Tt ) 
for this gene. More than 100,000 people in 
Italy have been classified as TT. Tt, or tt. 



Notice that in the case of thalassemia neither 
T nor / is completely dominant (nor com- 
pletely recessive). 

Although with respect to phenotypic ex- 
pression the relation between the alleles in 
the heterozygote may involve complete, 
partial, or no dominance, it should be re- 
called that gene action ordinarily has no 
effect upon either the integrity of the genes 
or their segregation and recombination. 

One can study a large number of other 
human traits by the pedigree and family 
methods and apply the principles known 
about genes to explain the data genetically, 
using the simplest suitable explanations in 
much the same way as was illustrated here 
for albinism and other traits. Sometimes, 
unfortunately, the data are insufficient and 
the investigator is left with several equally 
probable genetic explanations. 



SUMMARY AND CONCLUSIONS 

The gene is a unit or restricted portion of the total genetic material as discovered via 
any operational procedure. The genes discovered in the present chapter were revealed 
by recombination. 

Genes occur in pairs. When they are transmitted in sexual reproduction the mem- 
bers of a pair segregate so that any offspring receives only one member of a pair from 
each parent. The gene is uncontaminated by the type of gene that is its partner prior 
to segregation, and enters the new individual uninfluenced by the allele being con- 
tributed from the other parent. The hypothesis that the chromosomes comprise or 
carry the genetic material can be made more specific — a recombinationally detected 
gene may be represented by a short chromosome segment within which an exchange 
leading to a chiasma cannot occur. 

Data furnished in pedigree and family studies provide evidence that a number of 
human traits are based upon the action of a pair of segregating genes. 



REFERENCES 

Gates, R. R., Human Genetics, 2 Vols., New York: Macmillan, 1946. 

Mendel, G., 1866. "'Experiments in Plant Hybridization,*' translated in Sinnott, E. W., 
Dunn, L. C, and Dobzhansky, Th... Principles of Genetics, 5th Ed.. New York: 
McGraw-Hill, 1958. pp. 419-443; also in Dodson, E. O.. Genetics, the Modern 
Science of Heredity, Philadelphia: Saunders. 1956. pp. 285-311: also in Classic 
Papers in Genetics, Peters, J. A. (Ed.), Englewood Cliffs, N.J.: Prentice-Hall. 
1959, pp. 1-20. 



40 CHAPTER 3 

Mohr, O. 1 .. "Woolly Hair a Dominanl Mutanl c haractei in Man," J. Herod.. 23:345- 
352, 1932. 

Neel, J. V., and Schull, W. J., Human Heredity, University of Chicago Press, 1954, 
pp. 83 86, 89 91, 240 241. 

Stern. (.'.. Principles oj Human Genetics, 2nd Ed., San Francisco: Freeman, I960. 



QUESTIONS FOR DISCUSSION 

3.1. How would you recognize a line of garden peas that had become genotypically 
pure for a given trait'.' 

3.2. Criticize the assumption that genes come only from pre-existing genes and do 
not arise de novo. 

3.3. Differentiate between phenocopies and phenotypic overlaps. 

3.4. Does a parent lose its own genetic material when it is transmitted to progeny? 
Defend your answer. 

3.5. Is it necessary to assume that genes are able to reproduce themselves? Explain. 

3.6. List all the assumptions required to explain a 3:1 ratio in F 2 on a genetic basis. 

3.7. A mating of a black-coated with a white-coated guinea pig produces all black 
offspring. Two such offspring when mated produce mostly black but some white 
progeny. Explain these results genetically. 

3.8. A cross of two pink-flowered plants produces offspring whose flowers are red, 
pink, or white. Defining your genetic symbols, give all the different kinds of 
genotypes involved and the phenotypes they represent. 

3.9. What operation was employed in studying the gene in the present chapter? Define 
a gene in terms of size. 

3.10. Discuss the role of dominance in the study of genes. 

3.11. Do organisms that reproduce asexually have genes? Explain your answer. 

3.12. What relation has a gene to the phenotypic effect with which it is associated? 

3.13. Do you agree with the statement on p. 33 that a cross between two colorless 
pea plants results in "all colorless progeny"? Why? 

3.14. Throughout this book the use of the word "heredity" and its derivatives has 
been avoided. Why do you think this is, or is not, justified? 

3.15. What is the difference between the pedigree and family methods of investigation? 

3.16. What evidence is there that pigmentation (albinism vs. nonalbinism) is due to 
genes that are segregating? 

3.17. Two nonalbinos marry and have an albino child. What is the chance that the 
next child is albino? Nonalbino? That of the next two children, both are 
albinos? Nonalbinos? One albino and one nonalbino? 

3.18. What proportion of three-child families, whose parents are both heterozygous 
for albinism, have no albino children? All albino children? At least one albino 
child? 

3.19. Would you conclude that the gene for woolly hair is completely dominant to 
nonwoolly hair? Explain. 



Segregation of Alleles 41 

3.20. What are the similarities and differences regarding the segregation of genes and 
of chromosomes? 

3.21. If the genetic material is in the chromosome, is it necessary to assume that the 
members of a gene pair occupy exactly corresponding positions in the two homo- 
logs? Why? 

3.22. If a mitotic chromosome normally contains two identical chromatids, can we 
decide whether each chromosome contains one gene or an identical pair of genes? 
Justify your answer. 

3.23. In what respect do you suppose that a sample of two-child families may be 
biased? How would you attempt to avoid this error? 

3.24. The electron microscope shows that the sperm heads of some organisms contain 
a mass of uniformly thin threads. Do such cases offer evidence against the 
retention of the integrity of chromosomes or of genes? Why? 

3.25. Differentiate between the genetic recombinations that occur at the time of fertiliza- 
tion and at the time of meiosis. 

3.26. What is the difference between the study of genetics at the cellular and at the 
organismal levels? 

3.27. Describe the environment of a single gene. 

3.28. A lack of neuromuscular coordination, ataxia, occurs in the pedigrees of certain 
families in Sweden. How can you explain that one form of this rare anomaly 
occurs in certain families where the parents are apparently unrelated, and an- 
other form occurs in other families where the parents are first cousins? 

3.29. What bearing have the following facts relative to the generality of the phenomenon 
of dominance? 

When pure lines of smooth-seeded plants and shrunken-seeded plants are 
crossed, the Fj seeds are all smooth. Microscopic examination reveals that the 
margins of the starch granules in the seeds are smooth in the smooth P,, highly 
serrated or nicked in the shrunken P l9 and slightly serrated in all the F v 



Chapter 4 

INDEPENDENT RECOMBINATION 
BY NONALLELES 



I 



n i hi: preceding chapter we dis- 
cussed the transmission genetics 
of alternatives tor a single trait 
and found that a single pair of genes could 
explain the data in each case. But what is 
the genetic unit of transmission when two 
or more different traits are followed simul- 
taneously in breeding experiments? The 
answer may be found in the results of some 
additional experiments performed with the 
garden pea. 1 Other studies show that seed 
shape and seed color, like the flower color 
trait described in Chapter 3, are each due 
to a single pair of genes. That is, a P] of 
pure-line round X pure-line wrinkled seeds 
gives round F,, round being dominant. Self- 
fertilizing the F, round produces Fj in the 
proportion of 3 round: 1 wrinkled. Simi- 
larly, a P, of pure-line yellow x pure-line 
green seeds gives yellow F,, yellow being 
dominant, and self-fertilization of the yellow 
Fi gives 3 yellow: 1 green in F 2 . 

What actually happens in a crossing of 
individuals that differ simultaneously with 
regard to both seed shape and seed color? 
A round yellow strain is crossed with a 
wrinkled green strain, these strains being 
available as pure lines. In F, only round 
yellow seeds are obtained (Figure 4-1). 
This result is what we would expect had we 
been studying shape and color of seeds sep- 
arately. In this case, there is no phenotypic 

Based upon experiments of G. Mendel. 
42 



etleet ol the dominance of one trait upon the 
phenotypic expression o\ the other trait. 

Self-fertilization of the round yellow F, 
gives offspring which, when counted in sulli- 
ciently large numbers, occur approximately 
in the relative frequencies of 9 round \ el- 
low: 3 round green : 3 wrinkled yellow:! 
wrinkled green. Notice that segregation and 
recombination are involved for each trait, 
as revealed in F 2 by approximately 12 
round: 4 wrinkled and by about 12 yel- 
low^ green. In this generation also there 
is no effect of one trait upon the recombina- 
tional behavior of the genetic material for 
a different trait. 

From these results, what else can we de- 
cide regarding the gene? Until now, we 
have been able to explain all the experi- 
mental data on the basis of only a single 
pair of genes, as if the total genetic material 
of a diploid cell is divisible into only two 
genes, any one gene having numerous dif- 
ferent alleles, each one having effects on 
many different traits. To continue to con- 
sider that each Pi individual in the present 



Round Yellow x Wrinkled Green 

ALL Round Yellow 

F, Round Yellow x F, Round Yellow 

PHENOTYPE NUMBER RATIO 
Round Yellow 315 9.06 

Round Green 101 2.9 

Wrinkled Yellow 108 3.1 

Wrinkled Green 32 0.9 



FIGURE 4-1. Phenotypic results from studying 
two traits simultaneously. 



Independent Recombination by Nonalleles 



43 



case carries but a single pair of genes, each 
gene must have two simultaneous effects, 
one on seed shape and the other on seed 
color. The results obtained are consistent 
with this requirement in the following re- 
spect: the F, are round yellow, and the F L . 
give a 3 : 1 ratio for yellow vs. green and 
also for round vs. wrinkled. According to 
this hypothesis, the F L . should be of only two 
types — 3 round yellow: 1 wrinkled green. 
But in F L . not only are these grandparental 
(Pi) combinations found, but two new, re- 
combinational classes of offspring appear. 
namely, round green and wrinkled yellow! 
Apparently, then, what is genetically trans- 
mitted is not composed of a single pair of 
indivisible units, but is composed of pairs 
of units, or genes, with each gene pair ca- 
pable of undergoing segregation separately . 
Let us assume, therefore, that each sex- 
ually reproducing organism contains more 
than one pair of genes. In the present case, 
let/? (round) and r (wrinkled) be the alleles 
of one pair of genes and Y (yellow) and y 
(green) be the alleles of the second pair. 
The Pj, then, would be RR YY (round yel- 



RR YY 

RR Yy A n 

^ Round 

Rr YY W Yellow 
Rr Yy 



/2 Y 



RR yy 



r yy 



Round 
Green 



rr YY 



r r Y y 



Wrinkled 
Yellow 



r yy » 



Wrinkled 



figure 4-2. Expected genotypes and pheno- 
types in F 2 following segregation. 



'/a R 




— %RY 



— Va R y 



V 2 r 




y 2 y 



Va r Y 



y 2 y 



Va r y 



i IGl rh 4-3. Genotypes of gametes formed by 
a dihybrid, Rr Yy, undergoing independent 
segregation. 



low) and rr yy (wrinkled green). Each 
pair of genes would undergo segregation so 
that a gamete would contain only one mem- 
ber of each pair. In this manner the former 
parent would produce only RY gametes and 
the latter only ry, and all F, would be Rr Yy 
(round yellow), as observed. 

Based on the current hypothesis, the gam- 
etes formed by the Fj would contain either 
R or r. and. moreover, would contain either 
Y or y. Since the F 2 show that R and Y do 
not always go together into a gamete, nor 
do r and y, there must be four genotypes 
possible in gametes. RY. Ry. rY. ry. Since 
these possible haploid genotypes will be 
found both in male and in female gametes, 
it is expected that the ¥_. would contain the 
diploid genotypes and their corresponding 
phenotypes indicated in Figure 4—2. Note 
that nine different genotypes are possible in 
F 2 , four giving the round yellow phenotype, 
two giving round green, two wrinkled yellow, 






.--•••:• - 



C = = 5-R SG 



, 



. • • 



• • 



v w 



- • • 



- I 



/ 

- I 



rY 




: 



::0 



■ 


r: • . 


■ 


RR Yv 




Rr YY 




RxYy 




r r • 




r: 


- 


Rr Yy 


1 





Rt YY 
RrYy 

- YY 

- Yy 



RR YY 
RR Yy 
Rr YY 

Rr yy 
rr YY 
rr Yy 

"■ yy 



~ ; - - ' . r ; 



9 Round 
Yellovs 

. * . . - - 

j •'•--•, e c 
Yellow 

1 Wrinkled 

C- r r ~ 



- " • 




RrYy 
Rryy 
rr Yy 

"ryy 



J .■*.""." " :' ." :' :' " 



: - - _"- 



±i: -. F. - 
Tks qmsssk 



s of ame pmr of gams 



r_r= --"- 
R, of 

- . - - ._ ' 

■J-.c 

- - - 



Y. Rm. rY, and rr. 



Independent leles 






The branching track in Figure 4—4 can 
be read from the top: 1 4 of female gametes 
are /?y and are fertilized : 4 of the time by 
/?y mal; (producing %$ of all off- 

spring a Ul? > > | of the time fertiliza- 
tion is by .R > male gamete 
all offspring are RR > :rom this origin), 
etc. Summing up like classes, the kinds 
and relative numbers of gene n i of 

phenotypes are obtained as shown in the 
figure. The observed ratio (Figure 4—1 

. client agreement with the expected one. 

The branching track may be used to ob- 
tain the 9:3:3:1 pbenotypic ratio more 
simp. . know that crossing together 

two identical monohybrids (beterozygotes 
for only one of all the gene pairs under con- 
sideration ) yields a 3:1 phenotypic ratio of 
dominant : rec . . trait If the recombina- 
tional activity of two pairs of genes is inde- 
pendent- both pairs being heterozygous and 
showing dominance, then the two indepen- 
dently produced 3 : 1 ratios may be combined 
in the progeny as shown in Figure 
This may be read: among the offspring, the 
"ich are round (because of segregation, 
random fertilization, and the dominance of 
R in the cross Rr ■ Rr will also be yellow 
2 4 of the time and gre. d die time 

cause of segregation, random fertilization, 
and the dominance of J' in the cross 
?. of all progerr. 11 be 

round yellow and s 1<5 round greet 



see, men, mat independent segrega- 
tion by two pairs of genes results in the 

' ' ' - ■ '-' - - - i - ; ' ■-.:.'. : ' ■-.'.'. 

In the present case, the F 3 dihybrid (hetero- 
zygote for two of all the gene pairs under 
consideration ) received bom R and Y from 
.:..-.-'-. ■'._■ 

HV_ • ' - ? -'-... . . ':- 

parent and rY from the other, the gametic 
recombinations would have been RY and 

. . - - - 

Regardless of how die genes enter the in- 
dividual, then, the dihybrid forms four, 
equally frequent, genetically different gam- 

The types and frequencies of gametes 

: "' -- '. "-'- '■ : --'-. ..- - . ..- 

• . . , ~ - . . - .-----.--. • . . -. - - 

double recessive individual, Le.. an individ- 
ual homozygous recessive for both pairs of 
genes concerned. In the cross of Rr Yy by 
----- . :-.-:--. - 
- - . -. - . - . ::'--/-. - _ 

pected to produce four dif fe r e nt and equally 
:: "-'-.'- ■- -' R '':' \ -.::'-:- 

r - r - ~ '- *--"-; : f s r r. - 1 ± 

cross (Figure 4—6) that very nearly 1 

----- " " :-- If -_-: 

r . . f r 25 . _ 

"1 ' " ' - ---..:. ---- .:--•; ;- 



PARENTS Rr Yy x Rr Yy 



.; Round 



<- I ellow 



OFFSPRING < 



I '/« Wrinklec <^ 



Greei 

• - - 

Gfeei 



lound elow 

Rc-= G-ee- 
•Vrinlcled Yellow 
•'•-■ e r 3 '£ £ - 



FlGU? I — S Phenotypic results of a cross between identical dihybrids. 



16 



CHAPTER A 



GAMETES 



Va r y 



1 r y 



GENOTYPES 



Va rr yy 



PHENOTYPES 



9 


o* 






Va R Y 


1 ry 


'/< Rr Yy 


Va Round Yellow 


■ARy 


1 ry k 


V 4 Rr yy 


k Va Round Green 


'4 r Y 


1 ry 


r 'A rr Yy 


f Va Wrinkled Yellow 



'A Wrinkled Green 



figure 4-6. Test cross of the F 3 dihybrid ( Rr Yy) with the double recessive indi- 
vidual ( rr yy). 



of segregation by the members of a single 
pair of genes and of independent segrega- 
tion by different pairs of genes. 

Whenever one is dealing with complete 
dominance, a cross to an individual reces- 
sive for the pairs of genes involved will al- 
ways serve to identify the genotype of the 
other parent, since the phenotypic types and 
frequencies of the offspring will correspond 
to the genotypic types and frequencies oc- 
curring in the gametes of the latter. This 
kind of cross is, therefore, called a test cross, 
or a backcross when one of the parents in 
the series of crosses is homozygous recessive 
for the genes under study. 

We are now in a position to return to a 
consideration of the material basis for genes. 
If one gene pair is to be associated physically 
with the corresponding short regions in a 
pair of homologous chromosomes, within 
which an exchange leading to a chiasma 
cannot occur, the question is, where, in rela- 
tion to one pair of genes, is a second pair 
located? Two possibilities occur — either 
both pairs are on the same chromosome pair 
or they are on different, nonhomologous 
chromosome pairs. Consider the latter as- 
sumption — that different pairs of genes are 
located on different pairs of chromosomes. 
If this is true, then there are several differ- 



ent arrangements that the parts of different 
pairs of chromosomes may take relative to 
each other at metaphase I of meiosis (Fig- 
ure 4-7). 

It has been established that different pairs 
of chromosomes arrive at metaphase I in- 
dependently of each other. Moreover, it is 
entirely reasonable that the orientation to- 
ward the poles, of the centromeres in tetrads 
at metaphase I and in dyads at metaphase 
II, is not influenced by the presence or ab- 
sence of chiasmata or exchanges. If, as in 
Case A (Figure 4-7), no exchange occurs — 
and, hence, no chiasma is formed — between 
the centromere and gene pair A a or between 
the centromere and gene pair Bb, alignments 
I and II, being equally frequent, will result 
in four different, equally frequent types of 
gametes at the end of meiosis. The same 
result is also obtained either when there is 
a chiasma between the centromere and the 
gene in question in one tetrad but not the 
other (Case B), or when a chiasma occurs 
in each of the tetrads (Case C). Both in 
Case CI and CII the dyads can orient to 
the poles at metaphase II in four equally 
likely arrangements, with the same net re- 
sult, four equally frequent types of gametes. 
Therefore, independent segregation of dif- 
ferent pairs of chromosomes can serve as 



Independent Recombination by Nonalleles 



47 



Pole—— Metaphase I— ■- Pol< 



CASE A 
No chiasma 



II 



CASE B 
After one 
chiasma "^ 

in one pair 



or 



II 



Haploid Meiotic 
Products at 
Telophase II 



AB, AB, ab, ab 



Ab, Ab, aB, aB 



AB, Ab, aB, ab 



Ab, AB, ab, aB 



CASE C AB, ab, AB, ab 

After one J { \\ orA b, aB ' AB, ab 

chiasma fl I or A B, ab, Ab, aB 

in each pair or Bl Ub Bl Ub orAb, aB, Ab, aB 

Ab, aB, Ab, aB 
orAB, ab, Ab, aB 
orAb, aB, AB, ab 
orAB, ab, AB, ab 

figure 4-7. Meiotic fate of gene pairs presumably located in nonhomologous pairs of 
chromosomes. Note that when all alternatives in Case CI (and 11) are considered 
AB - ab Ab = aB. 



48 



CHAPTER 4 



Pole-* Metaphase I 



Pole 



Haploid Meiotic 
Products at 
Telophase II 



CASE A 

No chiasma 
between 
gene pairs 



n 



A a 



B b 



AB, AB, ab, ab 



CASE B 

After one 
chiasma 
between 
gene pairs 



It 



A a 
b B 



I 



AB, Ab, aB, ab 



figure 4-8. Meiotic fate of gene pairs presumably located in the same pair of 
chromosomes. 



the physical basis for independent segrega- 
tion of different pairs of genes, regardless of 
chiasma formation. 

Let us examine next the consequences of 
assuming that A and B are on the same chro- 
mosome, and a and b are on the homologous 
chromosome of the pair (Figure 4-8). 
When no chiasma occurs between the two 
different pairs of genes, Case A, only the 
old (maternal and paternal) combinations 
are found in the gametes. When such a 
chiasma occurs. Case B, four gametic classes 
are produced with equal frequency (two old 
and two new combinational types ) . But, un- 
less every tetrad has a chiasma in the region 
between the nonalleles, the number of old 
gene combinations found among the gametes 
will exceed the new combinations. Al- 
though a tetrad usually contains one or 
several chiasmata, there are numerous points 
along the chromosome where an exchange 



leading to a chiasma might occur. An addi- 
tional hypothesis would be needed if each 
tetrad were required to form a chiasma 
within a given interval, such as between A 
and B. Moreover, we have no knowledge 
as to the genie interval, or the distance be- 
tween nonalleles presumed to be in the same 
chromosome. Accordingly, we shall neg- 
lect, for the time being, the possibility that 
nonalleles in the same chromosome pair 
can form old and new combinations with 
equal frequency — that is, we shall assume 
that two pairs of genes which do so, and are 
therefore segregating independently of each 
other, are located in different pairs of chro- 
mosomes. Evidence consistent with this as- 
sumption is obtained from studies with gar- 
den peas. From the breeding behavior of 
hybrids for two or more gene pairs, it is 
possible to establish the existence of seven 
different pairs of genes (each happening to 



Independent Recombination by Nonalleles 



49 



AA' x AA' BB' x BB' AA' BB' x AA' BB' x V» 



Va AA 



Vi AA' 



figure 4-9. Genotypic re- 
combination frequencies. 



y 4 a'a' 



Va BB 


1 AA BB 


(1) 


Vi BB' 


\ 2 AA BB' 


(2) 


Va B'B' 


/ 

1 AA B'B' 


(3) 


Va BB 


2 AA' BB 


(4) 


Vi BB' 


\ 4 AA' BB' 
/ 


(5) 


Va B'B' 


2 AA' B'B' 


(6) 


Va BB 


1 A'A' BB 


(7) 


y 2 bb' 


\ 2 A'A' BB' 


(8) 


V4 B'B' 


f 

1 A'A' B'B' 


(9) 



show dominance in the hybrid condition), 
each pair seeming to segregate independently 
of all the others. Since the garden pea pos- 
sesses a diploid number of seven pairs of 
chromosomes, there are enough chromo- 
some pairs for each pair of genes to be lo- 
cated on a different pair of chromosomes. 

* Different Phenotypic Ratios - 

A monohybrid may show the phenotypic 
effects of only one allele, some of the effects 
of both alleles, or the complete effects of 
both alleles. These phenotypic conse- 
quences have already been designated as 
complete, partial, and no dominance, re- 
spectively. In the garden pea experiments 
already discussed, complete dominance pro- 
duced the 3 : 1 phenotypic ratio obtained 
from a cross between identical monohybrids. 
This necessitated breeding the offspring with 
the dominant phenotype in order to identify 
the 1:2:1 genotypic ratio predicted from 
such crosses. Had no dominance or partial 

-See W. Bateson (1909). 



dominance occurred, the phenotypic and 
genotypic ratios would have been identical. 
Nevertheless, in all cases the recombining 
genes retained their individuality, and the 
specific ratios observed depended only upon 
the dominance relation within the gene pair 
— that is, the relation between the expres- 
sion of one allele and that of its partner. 

Complete dominance also has no influence 
upon the individuality or segregation of non- 
allelic pairs of genes. Although the geno- 
typic ratio expected from crossing two par- 
ticular dihybrids has already been derived 
(Figure 4-4), let us rederive this ratio, em- 
ploying more general symbols for the genes, 
using the branching-track method in a 
slightly different way. Let A and A' be one 
pair of alleles and B and B' another. Mating 
AA' BB' by AA' BB' gives the genotypic 
ratio seen in Figure 4-9. 

Notice here that among every 16 off- 
spring, on the average, there would be 9 
different genotypes in the ratio of 1:2:1:2: 
4:2:1:2:1. How did this genotypic ratio 



50 



CHAPTER 4 



give rise to the 9:3:3: I phenotypic ratio in 
crosses between identically dihybrid garden 

peas'.' I'wo factors were responsible. One 
was the occurrence of dominance for each 

pair oi alleles; this converted the 1:2:1 
genotypic ratio for each gene pair to a 3:1 
phenotypic ratio. The other factor was that 
the action of the two gene pairs was inde- 
pendent and resulted in detectable effects on 
different traits. This permitted both 3:1 
ratios to be recognized separately even when 
these ratios were distributed at random in 
the progeny (Figure 4-5). It becomes ap- 
parent, therefore, that the particular pheno- 
typic ratios obtained, following crosses in- 
volving more than one gene pair, depend 
both upon the dominance relationships be- 
tween alleles and the gene interaction rela- 
tionships between nonalleles. 

If neither gene pair shows dominance, and 
if each pair acts both independently and on 
different traits, two 1:2:1 phenotypic ratios 
will be produced, and these, when distributed 
at random, will result in the 1:2:1:2:4:2: 
1:2:1 phenotypic ratio. Here, because no 



genotype is masked phenotypically by any 
other, the phenotypic and genotypic ratios 
are the same. (This would also be true of 
the following crosses: A A' BB' X AA BB, 
AA' BB' X A' A' B'B', AA' BB X AA BB'. ) 
This kind o\ result is illustrated in the prog- 
eny of parents both of whom have thalas- 
semia minor (77) and MN (MM') "blood 
type." (MM is phenotypically M, M'M' is 
phenotypically N, as described on p. 58.) 

When, however, the aforementioned con- 
ditions are changed so that one of the two 
pairs of genes shows dominance, two dif- 
ferent genotypes will produce the same 
phenotype, and fewer than 9 phenotypes are 
expected. Thus, if B is dominant to B\ 
genotypes 1 and 2 (in Figure 4-9) are ex- 
pressed as one phenotype, genotypes 4 and 
5 as another, and 7 and 8 as another, so that 
the phenotypic ratio becomes 3:1:6:2:3:1. 
This is the phenotypic ratio expected in the 
progeny of parents both heterozygous for 
albinism (Aa) and having MN blood type 
(MM'). If both gene pairs show domi- 
nance, one phenotype is expressed by geno- 




FIGURE 4-10. Drosophila melanogaster itui- 
tants showing the no-wing (left) and the curled 
wim; ( right ) phenotypes. ( Drawn by E. M. 
Wallace.) 



Independent Recombination by Nonalleles 



51 



types 1, 2, 4, 5, another by genotypes 3 and 
6, another by 7 and 8. and another by geno- 
type 9, producing the 9:3:3:1 ratio already 
discussed. Dominance causes a reduction 
in the number of phenotypic classes. 

What phenotypic ratios are expected when 
two different, independently active, pairs of 
genes affect the same trait in the same man- 
ner or direction? If one or more allelic 
combinations for one gene pair produce the 
same phenotype as one or more allelic com- 
binations for the other gene pair, the num- 
ber of phenotypes will also be reduced from 
the maximum (9 when identical dihybrids 
are crossed). Of course, the number of 
different phenotypes detected will be further 
reduced if the absence of dominance in both 
gene pairs is changed to dominance in one 
gene pair, and still further reduced if both 
gene pairs show dominance. Thus, if A 
and B produce equal amounts of melanin 
pigment in human skin, the amount of pig- 
ment being cumulative, and A' and B' pro- 
duce none, dominance being absent, a cross 
between identical dihybrids yields the ratio 
1 "black" (type 1 ) :4 "dark" (types 2, 4) : 
6 "mulatto" (types 3, 5, 7) :4 "light" (types 
6, 8):1 "white" (type 9), instead of 9 dif- 
ferent phenotypes. Moreover, if both A 
and B show complete dominance, for exam- 
ple either gene producing full flower color, 
the phenotypic ratio becomes 15 colored 
(types 1-8) :1 colorless (type 9). Note 
that when different pairs of genes act on 
the same trait in the same direction or way, 
they have a common phenotypic background 
on which their effects superpose, and the 
effect of one gene pair interferes with the 
detection of the effect of the other pair. 

Sometimes different gene pairs act inde- 
pendently on the same trait in different — 
antagonistic or cooperative — ways. For ex- 
ample, in Drosophila (Figure 4-10), A' is 
a recessive allele which reduces the wing to 
a stump, whereas B' is a recessive allele 



which causes the wing to be curled, the 
dominant allele A making for normal sized 
wings and the dominant allele B straight 
wings. A cross between two identical dihy- 
brids does not produce the customary 9:3: 
3:1 ratio. In the present case, the ratio 
becomes 9 flies with long, straight wings: 3 
with long, curled wings: 4 whose wings are 
mere stumps (of which one quarter would 
have had curled wings if the full wing had 
formed). Here, then, the phenotypic ex- 
pression of one gene pair can prevent de- 
tection of the phenotypic expression of an- 
other gene pair. 

In another case, either of two pairs of 
genes may prevent a given phenotype from 
occurring. Suppose the dominant alleles A 
and B each independently contribute some- 
thing different but essential for the produc- 
tion of red pigment, whereas their corre- 
sponding recessive alleles A' and B' fail to 
make the respective independent contribu- 
tions to red pigment production. Then 
crosses between two identical dihybrids will 
produce 9 red: 7 nonred (composed of 3 
homozygotes for A' only, 3 homozygotes for 
B' only, and 1 homozygote for both A' and 
B ') . Notice that if the recessive alleles are 
considered, we have just dealt with examples 
of unilateral and mutual opposition to pheno- 
typic expression, respectively, but if the dom- 
inant alleles are considered these become 
cases of unilateral and mutual cooperation 
in phenotypic expression. 

In all cases where two pairs of genes af- 
fecting the same trait interact phenotypically 
by superposition, antagonism or cooperation, 
one pair of genes has had an influence upon 
distinguishing the effects of the other. The 
general term epistasis may be used in these 
cases to describe the interference with — 
suppression or masking of — the phenotypic 
expression of one pair of genes by the mem- 
bers of a different pair. Genes whose de- 
tection is hampered by nonallelic genes are 



52 



CHAPTER 4 



said to be hypostatic, or to exhibit hypo- 
stasis. As dominance implies recessiveness, 

so epistasis implies hypostasis. There need 
be no relationship between the dominance 
o\ a gene to its allele and the ability of the 
gene to be epistatic to nonalleles. In theory. 
then, epistatic action may depend upon the 
presence of A. A\ or A A'; moreover, hypo- 
static reactions may depend upon the pres- 
ence of B, B\ or BB'. Consequently, it 
should be noted that in crosses between 
identical dihybrids. epistasis-hypostasis can 
produce phenotypic ratios still different from 
those already described. 

Consider another example of a dihybrid 
in which both pairs of genes show domi- 
nance but no epistasis. In Drosophila, the 
dull-red eye color of flies found in nature 
is due to the presence of both red and brown 
pigments. Let A be the allele which pro- 
duces the red pigment and A' its recessive 
allele which produces no red pigment; let B 
be the nonallele producing the brown pig- 
ment whose allele B' makes no brown pig- 
ment. A mating between two dull-red dihy- 
brid flies (from a cross of pure red, A A B'B' 
by pure brown. A' A' BB) produces offspring 
in the proportion 9 dull-red (containing 
A — B — ):3 red (containing A — B'B'): 3 
brown (containing A' A' B — ) : 1 white (A' A' 
B'B'). The last phenotypic class, resulting 
from the absence of both eye pigments, is 
new in this series of crosses. This case illus- 
trates that the interaction of nonallelic genes 



may result in apparently novel phenotypes. 
Such interactions change not the number but 
the kind of phenotypes obtained. 

The preceding discussion suggests that any 
given phenotypic trait may be the result of 
the interaction of several gene pairs. One 
is even led to conclude that the total pheno- 
type is the product of the total genotype 
acting together with the environment. The 
difference between phenotypic and geno- 
typic ratios is often due to products of gene 
action — by alleles and nonalleles — which 
superpose, cooperate, or conflict at the phys- 
iological or biochemical level. It is also pos- 
sible that there is sometimes a direct in- 
fluence of one gene upon the ability of an 
allele or nonallele to act. Although the na- 
ture of gene interactions can be predicted 
partially, in a general way. from the kind 
of modified ratio obtained, an understanding 
of the mechanisms involved must ultimately 
be based upon a knowledge of how genes 
act and the nature and fate of gene products. 
In no case has a phenotypic ratio that differs 
from the expected genotypic one served to 
disprove either segregation or independent 
segregation. In fact, segregation and inde- 
pendent segregation were first proved despite 
the misleading phenotypic simplifications of 
genotypic ratios wrought by the occurrence 
of dominance; moreover, the principle of 
independent segregation could also have been 
first proved from crosses involving epistasis 
or apparently novel phenotypes. 



SUMMARY AND CONCLUSIONS 

When two different traits were studied separately, the phenotypic alternatives were 
found to be due to the presence of a single pair of genes in each case. Studies were 
then made of the distribution of phenotypes in successive generations when these two 
pairs of traits were followed simultaneously in the same individuals. The data ob- 
tained showed that each trait is due to the presence of a different pair of genes, proving 
that the genetic material is made up not of a single segregating pair of genes but of a 
number of segregating gene pairs. Moreover, the results are best explained by the 
principle that the segregation of one pair of alleles is at random with respect to the 
segregation of all the other nonalleles tested. The simplest hypothesis for the physical 



Independent Recombination by Nonalleles 53 




William Bateson (1861-1926). 

(From Genetics, vol. 12, 

p. 1, 1927.) 



basis for the independent recombination of such nonalleles is that different pairs of 
genes reside in nonhomologous pairs of chromosomes. 

The phenotypic expression of genes depends upon their alleles, insofar as dominance 
is involved, and upon nonalleles, insofar as epistasis (including superposition, coopera- 
tion, and antagonism) and the production of apparently novel phcnotypes are involved. 
The absence both of dominance and of epistasis will always produce phenotypic ratios 
which directly represent genotypic ratios, whereas the occurrence of one, the other, 
or both reduces the number of phenotypic classes. In any case, segregation and inde- 
pendent segregation, being genetic properties, are totally uninfluenced by the manner 
whereby genes do or do not come to phenotypic expression. 



REFERENCES 

Bateson, W., Mendel's Principles of Heredity, Cambridge, England: Cambridge Uni- 
versity Press, 1909. 

Mendel, G. See reference at the end of Chapter 3. 

Supplement I (at the end of this book). 



QUESTIONS FOR DISCUSSION 

4.1. Make genetic diagrams for the crosses and progeny discussed in the second and 
third paragraphs on p. 42. Be sure to define your symbols. 

4.2. Is a test cross or backcross used to determine genotypes from phenotypes in cases 
of no dominance? Explain. 



~>l CHAPTER 4 

4.3. What types and frequencies of gametes are formed by the following genotypes, 
all gene pans segregating independently? 

(a) An lib C< (C) A a lib Cc 

(b) AABBCcDD (d) MmNnOoPp 

4.4. How mam different diploid genotypes are possible in offspring from crosses in 
which both parents are undergoing independent segregation for the following 
numbers o\ pairs of heterozygous genes — 1, 2, 3, 4. n? 

4.5. What conclusions could you reach about the parents if the offspring had pheno- 
types in the following proportions? 

(a) 3 : 1 (c) 9 : 3 : 3 : 1 

( b ) I : I ( d ) 1 : 1 : 1 : 1 

4.6. Would you be justified in concluding that a pair of chromosomes can contain 
only a single pair of genes? Explain. 

4.7. Suppose a particular garden pea plant is a septahybrid. What proportion of its 
gametes will carry all seven recessive nonalleles? All seven dominant nonalleles? 
Some dominant and some recessive nonalleles? 

4.8. What proportion of the offspring of the following crosses, involving independent 
segregation, will be completely homozygous? 

(a) AaBbXAaBb (c) Aa BB Cc X A A Bb cc 

(b) AA BBCCX AA bbec (d) AA' X A" A'" 

4.9. Following independent segregation, why would you expect that gametes fertilize 
at random with respect to their genotypes? 

4.10. Discuss the particulate nature of the genetic material. 

4.11. Does the discovery of independent segregation of nonalleles affect your concept 
of gene size? Explain. 

4.12. Discuss your current understanding of the term "genetic recombination." 

4.13. Discuss the factors that can modify the expected phenotypic ratio. 

4.14. In snapdragons, red flowers ( R ) are incompletely dominant to white (/), the 
hybrid being pink; narrow leaves (N) are incompletely dominant to broad leaves 
(/;), the hybrid being intermediate in width ("medium"). Assuming the gene 
pairs recombine independently, give the genotypic and phenotypic ratios ex- 
pected among the progeny of a cross between the following: 

(a) a red medium and a pink medium plant 

(b) a pink medium and a white narrow 

(c) two identical dihybrids 

4.15. Suppose an albino child also suffers from thalassemia minor. Give the most 
likely genotypes of the parents. 

4.16. How can you explain that a certain kind of baldness, due to a single gene pair, 
is dominant in men and recessive in women? 

4.17. Though blue-eyed couples ordinarily have only blue-eyed progeny, brown-eyed 
couples may also have blue-eyed children. Select and define gene symbols so 
you can give the complete genotypes of all individuals mentioned in each of 
the families listed: 

(a) One member of a pair of twin boys has brown eyes, the other has blue eyes. 

(b) A blue-eyed, woolly-haired child resembles his father in one of these respects 
and his mother in the other respect. 

(c) A brown-eyed, nonthalassemic child is like the grandmother but unlike the 
mother in both of these respects. 



Independent Recombination by Nonalleles 



55 



4.18. Differentiate between dominance and epistasis. 

4.19. What is the maximum number of genotypes possible in the progeny if the parents 
are monohybrids? 

4.20. Two green corn plants are crossed and produce offspring of which approximately 
'■'\,; are green and 7 ic are white. How can you explain these results? 

4.21. Does gene interaction occur only when identical monohybrids (or identical di- 
hybrids) are crossed? Explain. 



.MA. 






WALNUT COMB 




SINGLE COMB 



4.22. A chicken from a pure line of "rose" combs is mated with another individual 
from a pure line of "pea" combs (see the accompanying illustration). All the 
Fj show "walnut" combs. Crosses of two F, "walnut" type individuals provide 
F 2 in the ratio 9 "walnut": 3 "rose": 3 "pea":l "single." Choose and define gene 
symbols to provide a genetic explanation for these results. 

4.23. Three walnut-combed chickens were crossed to single-combed individuals. In 
one case the progeny were all walnut-combed. In another case one of the progeny- 
was single-combed. In the third case the progeny were either walnut-combed or 
pea-combed. Give the genotypes of all parents and offspring mentioned. 

4.24. Matings between walnut-combed and rose-combed chickens gave 4 single, 5 pea. 
13 rose, and 12 walnut progeny in F v What are the most probable genotypes 
of the parents? 

4.25. A mating of two walnut-combed chickens produced the following F x with respect 
to combs: 1 walnut, 1 rose, 1 single. Give the genotypes of the parents. 

4.26. The hornless, or polled, condition in cattle is due to a completely dominant gene, 
P, normally horned cattle being pp. The gene for red color (R) shows no domi- 
nance to that for white (/?'), the hybrid (RR') being roan color. Assuming in- 
dependent segregation, give the genotypic and phenotypic expectations from the 
following matings: 

(a) Pp RRX pp RR' 

(b) Pp RR' X pp RR 

(c) Pp RR' X Pp RR' 

(d) hornless roan (whose mother was horned) X horned white 

4.27. When dogs from a brown pure line were mated to dogs from a white pure line 
all the numerous F] were white. When the progeny of numerous matings be- 
tween Fj whites were scored there were 118 white. 32 black, and 10 brown. 
How can you explain these results genetically? 

4.28. Using your answer to the preceding question, give the phenotypic and genotypic 
expectations from a mating between the following: 

(a) a black dog (one of whose parents was brown) and a brown dog 

(b) a black dog (one parent was brown, the other was black) and a white dog 
(one parent was brown, the other was from a pure white strain) 



56 CHAPTER 4 

4.2 l >. When one crosses pure White 1 eghorn poultry with pure White Silkies, all the 
I are while. In the 1 ■'■_.. however, large numbers of progeny occur in a ratio 
approaching 13 white:3 colored. Choosing and defining your own gene symbols, 
explain these results genetically. 

4.30. (a) In the yellow daisy the Bowers typically have purple centers. A yellow- 
centered mutant was discovered which when crossed to the purple-centered type 
gave all purple-centered I,, and among the F 2 47 purple and 13 yellow. Explain 
these results genetically . 

(b) later, another yellow-centered mutant occurred which also gave all purple 
F, from crosses with purple-centered daisies. When these F, were crossed to- 
gether, however, there were 97 purples and 68 yellows. Explain these results 
genetically. 

(c) How can you explain that a cross between the two yellow-centered mutants 
produced all purple-centered F,? 

4.31. Give a single genetic explanation that applies to all the following facts regard- 
ing human beings: 

(a) One particular deaf couple has only normal progeny. 

( b ) One particular deaf couple has only deaf progeny. 

(c) One particular normal couple has many children, about % are normal and 
Y 4 deaf. 

(d) One particular normal couple has all normal children. 

(e) Normal, identical twins marry normal, identical twins and have a total of 
9 normal and 9 deaf children. 

4.32. How can you explain the observations with regard to lint color of cotton that 
brown X green gives green F,, which when mated together produce F-, which 
contain mostly brown, some greens, and a few whites? 

4.33. Suppose two unrelated albinos married and had 8 children, 4 albino and 4 non- 
albino. How could you explain these results? 

4.34. When, during the life cycle, can dominance and/or epistasis occur or not occur 
in maize? Neurospora? 

4.35. When two plants are crossed it is found that °%4 of the progeny are phenotypically 
like the parents, and 1 <u of the progeny are different from either parent but re- 
semble each other. Give a genetic explanation for this. 

4.36. Would you expect to find epistasis in man in marriages involving genetic alterna- 
tives for both woolly hair and baldness? Brown eyes and albinism? Baldness 
and brown eyes? 

4.37. Assume, in man, that the difference in skin color is due primarily to two pairs 
of genes which segregate independently: BB CC is black, bb cc is white, any three 
of the genes in black produce dark skin, any two medium skin, and any one 
produces light skin color. Give the genotypes of parents who are: 

(a) Both medium, but have one black and one white child. 

(b) Both black but have an albino child. 

(c) Both medium and can have only medium children. 

(d) Medium and light and have a large number of children: % medium, : }s light, 
's dark, \s white. 



Chapter 5 

MULTIPLE ALLELES; 
MULTIGENIC TRAITS 



A L 



ll the phenotypic results dis- 
cussed in preceding chapters 
can be explained genetically 
by dividing the genetic material into gene 
pairs. Since no proof has been presented 
that a particular gene can occur in more than 
two different states, one could maintain at 
this point that the only alternative for a given 
gene (causing round seeds) is the absence of 
that gene (thus causing wrinkled seeds). 
This may be called the presence-absence 
view of gene alternatives. It is clear that it 
requires finding a gene with more than two 
alternatives to prove that not all mutations 
remove an entire gene, and that genes can 
mutate to alternative gene forms. If, as as- 
sumed, genes arise only by gene replication, 
one would expect to find multiple alleles, 
since each individual carries many different 
nonallelic genes, many of which must have 
been derived from a common ancestral gene 
in past evolution. 

Multiple Alleles 

1. Human Blood Types. Numerous family 
studies of blood type provide us with data 
regarding the number of alternatives possible 
for an allele. However, before discussing 
these studies, it is necessary to learn what 
is meant by a blood type or blood group. 

Human blood contains red blood cor- 
puscles (cells) carried in a fluid medium, 
the plasma. The corpuscles carry on their 
surfaces substances called antigens, whereas 
57 



the plasma contains, or may form, substances 
called antibodies. An antibody is a very 
specific kind of molecule capable of reacting 
with and binding a specific antigen. This 
reaction may be visualized as a lock (anti- 
body) which holds or binds a particular key 
(antigen). If a rabbit is injected with suit- 
able antigenic material — foreign red blood 
cells, for example — to which it has never 
before been exposed, certain antibody-pro- 
ducing cells of the rabbit will manufacture 
an abundance of antibodies, which will ap- 
pear in its plasma. Some of these antibodies 
will be used to react specifically with the 
antigenic component of the foreign red blood 
cells. If, on some later occasion, the same 
antigen is injected into the rabbit's blood- 
stream, specific antibodies will already be 
present to bind the antigen. The antigen- 
antibody complex then formed often causes 
the blood to clump, or agglutinate. It is 
simple to arrange the procedure so that this 
reaction may be observed in a test tube or 
on a glass slide. 

Red blood corpuscles from different peo- 
ple are injected into different rabbits, with 
the result that the rabbits form antibodies 
against the antigens introduced. The iso- 
lated rabbit's blood, centrifuged free of cells, 
can then serve as an antiserum, containing 
antibodies that will clump any red blood cells 
added to it carrying the original types of 
antigens. It is found 1 that two very dis- 
tinct antisera are formed by these rabbits, 
and that any person's blood cells tested with 
these two antisera can react in one of three 
ways: the red blood cells are agglutinated 
or clumped either in one antiserum (arbi- 
trarily called anti-M), or in the other (called 
anti-N), or in both of these antisera. All 
people can be classified by their blood cell 
antigens as belonging to either M, or N, or 
MN blood group, respectively. 

1 Based upon work of K. Landsteiner and P. 
Levine. 



58 



( HAPTl K 5 





PARENTS 






CHILDREN 












M 


MN 


N 


1 


M 


X 


M 


ALL 


- 


- 


? 


N 


* 


N 


- 


- 


ALL 


3 


M 


X 


N 


- 


ALL 


- 


4 


MN 


x 


N 


- 


Vi 


Vi 


5 


MN 


X 


M 


V, 


Vi 


- 


6 


MN 


x 


MN 


V* 


•A 


'/« 



FIGURE 5-1. Distribution of MN blood group 
phenotypes in different human families. 



Parents and their offspring can be tested 
for MN blood type. The results of such 
family studies are summarized in Figure 5-1. 
Parents of type 6 produce offspring in the 
proportion of 1:2:1 for M : MN : N blood 
types. This result suggests that these blood 
types are due to the action of a single pair 
of genes. If we let M represent the gene 
for blood antigen M, and M\ the allele which 
produces the N blood antigen, mating 6 must 
be, genetically. MM' x MM' and the off- 
spring \ MM:2MM':\ M'M' . Note that 
these alleles show no dominance. MM' indi- 
viduals having both M and N blood antigens. 
All the other family results also are con- 
sistent with the genetic explanation pro- 
posed. 

Different antisera can be prepared to test 
for other blood types. One of these anti- 
sera determines the presence or absence of 
what is called the Rhesus or Rh factor. Red 
blood cells from Rhesus monkeys are in- 
jected into rabbits; if a second injection of 
Rhesus blood is given sometime later, it will 
be clumped. This is explained by the pres- 
ence of an antigen carried by Rhesus red 
blood cells against which the rabbit had 
manufactured antibodies before its second 



exposure to Rhesus blood. The antigen in- 
voked here is called Rh: the antibodies in- 
duced are anti-Rh. 

When human blood is injected into a rab- 
bit having anti-Rh antibodies in its serum, 
it is found that 859? of all people have blood 
which is clumped — these people have what 
is called Rhesus-positive (or Rh-positive) 
blood type; 15% of all people have blood 
which is not clumped — these people have 
Rhesus-negative (or Rh-negative) blood 
type. Accordingly, 85 r /< of human beings 
have the same Rh antigen as have Rhesus 
monkeys, and 15% do not. A combination 
of family and pedigree studies shows that 
presence of Rh antigen in human beings is 
controlled by a dominant gene we can repre- 
sent by R, and its absence by a recessive 
allele, r. 

Two other antisera, called anti-A and anti- 
B, can be prepared.- Blood from different 
people tested with these antisera is found 
to be of one of four types: clumped in anti-A 
(blood type A), clumped in anti-B (blood 
type B), clumped in both (type AB), and 
clumped in neither type of antiserum (O). 

Family studies of ABO blood types give 
the phenotypic results shown in Figure 5-2. 
Note that two kinds of results are obtained 
from A X O and also from B X O parents. 
In each case one kind of result (marriage 
types 9 and 1 1 ) can be explained if one 
assumes that the non-O parent is a hetero- 
zygote in which the gene for O is recessive. 
Let i be the gene for O blood type and / ' 
the allele for A blood type, the latter being 
dominant. Then the parents are thus: in 
marriages type 9, /'/' X ii\ in type 10, 
/'/' X u\ and in 13, ii X "• In order to 
explain 11 and 12 we shall have to assume 
the presence of a gene /'•' for B blood type, 
which is also a dominant allele of i and from 
which it segregates. Then mating 1 1 is 

- Based upon K. Landsteiner's work. 



Multiple Alleles; Mtdtigenic Traits 



59 



I B i X U and 12 is I "I" X w. Note that wc 
have made a new supposition with regard 
to genes. In the former case the alternative 
allelic form of / is /', whereas in the latter 
case it is / /; . This must mean that / ' and / /; 
are also alleles. The results of marriage 
types 7 and 8 confirm this hypothesis, the 
heterozygote I A I B showing no dominance and 
appearing as AB blood type. All the results 
indicated in the table are now explained ge- 
netically. Note that l A and /" produce qual- 
itatively different antigenic effects. We 
have, therefore, proved that the gene can 
exist in any one of two or more alternative 
genetic states, so that a gene can have mul- 
tiple — different — alleles. Of course, any one 
person normally carries no more than two 
of all the multiple alleles possible. 

2. Blood Type Isoalleles. It has been 
shown that persons with A blood type really 
have one of three different subtypes, result- 
ing from slightly different allelic forms of 
/J __/-«, ia2 ja3 Three s iig ht i y different 



allelic alternatives are known also for / /{ , 
producing three subtypes within the B blood 
group. Thus, alleles which at first seem 
identical may prove to be different when 
tested further. Such alleles are said to be 
isoalleles. Other examples of isoalleles have 
been detected because different alleles show 
varied responses to the presence of non- 
allelic genes, to environmental changes such 
as temperature and humidity, or to agents 
which modify mutation rates. Of course, 
the number of isoalleles detected will de- 
pend upon how many different phenotypic 
criteria are employed to compare alleles, 
and how small a phenotypic difference is per- 
ceptible. 

In the case of ABO blood type, it is 
sometimes adequate to classify individuals 
on the basis of alleles that produce A-type, 
B-type, or neither type of antigen. There- 
fore, only three alleles need to be considered. 
When one studies the genetic relationships 
among individuals in detail, however, it is 
often necessary to deal with all seven alleles. 





PARENTS 






CHILDREN 












A 


AB 


B 





7 


AB 


X 


AB 


'A 


V* 


'A 


— 


8 


AB 


X 


O 


'A 


— 


Vi 


- 


9* 


A 


X 


O 


Vj 


— 


- 


'A 


10' 


A 


X 


o 


ALL 


— 


— 


- 


11* 


B 


X 


O 


- 


- 


Vj 


Vi 


12* 


B 


X 





- 


- 


ALL 


- 


13 





X 


O 








ALI 



figure 5-2. Distribution of ABO blood group 
phenotypes in different human families. 
* In some families. 



3. Isoalleles in Drosophila. In different 
wild Drosophila populations, designated as 
1, 2. and 3, the venation of the wings is 
complete and identical. In the hybrids pro- 
duced by all possible crosses between these 
populations, the venation is unchanged. 
This result suggests that all three popula- 
tions are genotypically identical in this re- 
spect. The venation in a mutant strain is 
incomplete, the cubitus vein being inter- 
rupted (ci = cubitus interruptus) in homo- 
zygotes (Figure 5-3). Hybrids formed by 
crosses between ci ci and wild populations 
1 or 2 have complete venation, so that the 
gene for normal venation, ci + , in these pop- 
ulations is completely dominant to ci. But 
the hybrid between ci ci and wild flies from 
population 3, c7 + :! ci, shows the cubitus vein 
interrupted. Furthermore, the lack of domi- 
nance of ci + : '' over ci can be shown to be 



60 



CHAPTER 5 





figure 5-3. Normal (a) and cubitus inter- 
rupts (/>) wings of Drosophila melanogaster. 
(Courtesy of C. Stern; from Genetics, vol. 28, 
p. 443, 1943.) 



an effect of this gene pair rather than a 
modifying effect of some other gene pair. 
Apparently, then, the ci + allele in popula- 
tion 3 is different from that in populations 
1 and 2. We are dealing, consequently. 
with two isoalleles in a multiple allelic series. 
(Note that a slightly different system of 
symbolizing genes was used here.) 

4. Eye Color in Drosophila. Another series 
of multiple alleles in Drosophila involves eye 
color. In this case the different alleles can 
be arranged in a series that shows different 
grades of effect on eye color, ranging from 
dull-red to white: dull-red (w + ), blood 
(u'-'). coral (W"), apricot (u"'). buff (w 6/ ), 
and white (w). The w * allele is dominant 
to the others listed and is the allele com- 
monly found in wild-type flies. One can 
think of all the different alleles as producing 
the same kind of phenotypic effect, but less 
of it in proceeding from w + to w, the white 
allele being completely inefficient in this 
respect. 

We have already described isoalleles for 
genes normally expressed in individuals liv- 
ing in the wild (wild-type isoalleles). Iso- 
alleles for mutant genes (mutant isoalleles) 
also occur. For instance, it has been proved 



that the gene producing white eye color in 
different strains of Drosophila is actually 
composed of a series of multiple isoalleles 
(vv 1 , w'-, w :! , etc.). 

5. Self-sterility in Nicotiana. Among sex- 
ually reproducing plants it is not uncommon 
to find that self-fertilization does not occur 
even though the male and female gametes 
are produced at the same time on a given 
plant. The reason for this has been studied 
in the tobacco plant, Nicotiana, where it was 
found that if pollen grains fall on the stigma 
of the same plant, they fail to grow down 
the style. When this happens self-fertiliza- 
tion is impossible. A clue to an explanation 
for this phenomenon comes from the ob- 
servation that different percentages of pollen 
from a completely self-sterile plant may 
grow down the style of other plants. 

The results of certain crosses are shown 
in Figure 5-4. Genetically identical pistils 
are exposed to pollen from the same plant 
(A), from a second one (B), and from a 
third (C). No pollen, approximately half, 
and approximately all, respectively, are able 
to grow down the style of the host. Note, 
in B, that although all the pollen used came 
from one diploid individual, half of it can 



Multiple Alleles; Mucigenic Traits 



61 



and half of it cannot grow on its host. Re- 
call that the stigma and style are diploid 
tissues, whereas pollen grains are haploid. 
These results suggest that most important in 
determining whether or not a pollen grain 
can grow down a style is not the diploid 
genotype of its parent but the haploid geno- 
type contained in itself. 

Let us assume that self- or cross-sterility 
is due to a single pair of genes. Call s3 the 
allele contained in the pollen which permits 
pollen to grow in case B. The pollen grains 
from the host plant furnishing the pistil can- 
not contain s3, or the pollen would be able 
to grow on their own parent; and they can- 
not (Case A). So, the host pistil tissue in 
this experiment cannot contain s3, and one 
of its alleles can be called si. Then, half 
of the pollen from the host individual will 
carry si (Case A); but since these fail to 
grow, we must assume that any pollen grain 
carrying an s allele also present in the host 
pistil will fail to grow. Excluding the pos- 
sibility of a mutation, the other allele in the 



host pistil cannot also be 57, since one si 
would have had to be received from a pa- 
ternal pollen grain growing down a maternal 
style that carried si as one of its two alleles. 
Since the second allele in the pistils illus- 
trated cannot be either si or s3, let us call 
it s2. The other half of pollen from the 
pistil parent thus will contain s2, and also 
fail to grow in self-pollination (Case A). 
In B the pollen grains that fail to grow are 
either s/ or s2 (adhering to the law of par- 
simony) ; however, their precise identity can- 
not be determined without additional tests. 
In C, since all the pollen grew, one pollen 
allele must be a different one — call it s4. 
The other pollen allele may be s3 or a still 
different one, s5. Here again more tests are 
needed to determine the precise identity. 

In these cases the phenotypic alternatives 
for pollen are to grow or not to grow. 
Whenever the pollen grains from any one 
plant are placed on a given stigma and both 
alternatives occur, the phenotypes are in a 
1 : 1 ratio. All these results and others are 



Stigma 



VA *'% 



PISTIL < Style 



figure 5-4. Multiple alleles 
for cross- or self -sterility. 



Ovary 



S1S2 



STS2 



S1S7 



62 



CHAPTER 5 



consistent with the assumptions made, that 
self- 01 cross-sterility is regulated by a single 
pair of genes which form a multiple allelic 
series. Some species prose to have fifty or 
more multiple alleles forming a series re- 
sponsible for self-sterility, group sterility, or 
group incompatibility. 

^Multigenic Traits 

Up to now. the traits chosen to study genes 
occur in clear-cut. qualitatively different al- 
ternatives like flower color in garden peas, 
or albinism vs. pigmentation, and various 
blood types in human beings. These are 
called discontinuous or qualitative traits be- 
cause in each case an individual belongs 
clearly to one phenotypic class or another. 
Although the interaction of many or all genes 
may ultimately be involved in the appearance 
of a given phenotype, the phenotypic alter- 
natives previously considered have been ef- 
fected primarily by only one or a few pairs 
of genes. Moreover, in these cases the non- 
genetic environment had much less or no 
effect upon the phenotypic differences in- 
volved. 

For practical and theoretical reasons one 
may also be interested in the genetic basis 
for certain continuous traits like height of 
corn or intelligence in man, for which there 
are so many grades that individuals are not 
separable into discrete types or classes. 
Such traits are also called quantitative traits 
because the continuous range of phenotypes 
observed requires that an individual be meas- 
ured in some way in order to be classified. 
Are quantitative traits also determined ge- 
netically? Let us make the simplest assump- 
tion that quantitative traits differ from qual- 
itative ones only in degree, the former being 
due to the combined effects of many gene 
pairs. In the case of multigenic (polygenic) 
traits, although many phenotypic classes 
would be made possible by the action of 
multiple gene pairs, the effect of any single 



pair would be difficult to distinguish. Con- 
sequently, since each pair of genes would 
contribute only slightly toward the expres- 
sion of the quantitative trait, one would 
expect the effect of environment to be rela- 
tively larger than that of any single gene 
or gene pair. The large effect of fertilizer 
upon corn ear si/e and of diet upon height 
in human beings illustrate the importance of 
environment in multigenic traits. 

A given trait may be determined qualita- 
tively in certain respects and quantitatively 
in other respects. For example, in garden 
peas one pair of genes may determine 
whether the plant will be normal or dwarf, 
the actual size of a normal plant being de- 
termined by multigenic interaction with the 
environment playing a significant role. Sim- 
ilarly, a single pair of genes can determine 
whether a human being has a serious mental 
deficiency or normal mentality, though nearly 
all individuals have a degree of mental ability 
which varies in a continuous way due to 
environment and polygenes. 

If quantitative traits are determined multi- 
genically, it ought to be possible to derive 
other characteristics of them which are con- 
sistent with actual observations by consider- 
ing the same trait, first as a qualitative trait 
(i.e., determined by one or two or three 
gene pairs), and then as a quantitative trait 
(i.e., determined by many gene pairs). Let 
the trait be color, and the alternatives in P, 
be black and white. Assume first that there 
is no dominance or epistasis (see p. 51); 
then, whether one, two, three, or many gene 
pairs are involved, the F, will be uniform 
and phenotypically intermediate (medium 
gray) between the two P,. Examine, in 
Figure 5-5, results of matings between F] 
(by cross- or self-fertilization) in each case. 
As the number of gene pairs increases, the 
number of classes of Fj offspring increases. 
As the number of classes becomes large, one 
would expect environmental action to cause 



Multiple Alleles; Multigenic Traits, 



63 



individuals to fall out of their phcnotypic 
class, so to speak, and into the space be- 
tween classes or into an adjacent phenotypic 
class. And so, as gene pair number in- 
creases, classes become more numerous, then 
indiscrete, resulting finally in a continuum 
of phenotypes. 

Note also that as the number of gene pairs 
determining the trait increases, the fraction 
of all F 2 resembling either P, becomes 
smaller. Thus, with one pair of genes y 2 
of F L > are black or white, with two pairs %, 
with three pairs Y S2 , etc. Consequently, as 
the number of genes increases from 1 to 20 
and more, the continuous distribution of 
phenotypic types gives rise to an F 2 curve 
which becomes narrower and narrower. In 
other words, the chance of recovering in F 2 
any phenotype a given distance off the mean 
decreases as gene pair number increases. 
Although it may be relatively easy to identify 
whether one, two, or three gene pairs cause 
a given characteristic, it is much more diffi- 
cult to determine exactly how many pairs 
are involved whenever more than three are 
involved. In multigenic cases, measurement 
of how the population varies relative to the 
average phenotype can give information as 
to the approximate number of polygenes 
involved. 

The variability of a trait can be measured 
statistically as follows: the mean, m (the 
simple arithmetic average), is found. The 
variance, v (the measure of variability from 
the mean), for a group of measurements is 
determined by finding the difference between 
each measurement and the mean, squaring 
each such difference, adding all the values 
obtained, and dividing the total by 1 less 
than the number of measurements involved. 
With a given sample size, all other things 
being equal, the greater the variance the 
smaller the number of gene pairs involved, 
as would be expected from Figure 5-5. One 
may find detailed statistical procedures for 



P x P, 



1 



F - >< F , 



F, < 




involving one gene pair 




involving two gene pairs 



-■ill.. 

involving three gene pairs 




involving many gene pairs 



figure 5-5. Dependence of number of pheno- 
typic classes upon number of gene pairs. 
Horizontal axis shows classes, vertical axis 
indicates relative frequencies. 



64 



CHAPTER 5 



3 /4 A- 



% B 



Va bb 



9 /i6 A- B- (1 unit) 



V, 6 A- bb (2 units) 



V4 a a 



% B- 



y 4 bb 



_^ 3 /i6 aa B- (0 units) 



.^ Vi6 aa bb (1 unit) 



figure 5-6. Results of cross- 
ing together the dihybrids 

described in the text. 



using variance this way in any standard text 
on elementary statistical methods. 

Consider next the effect of dominance 
upon the expression of quantitative traits. 
When a qualitative trait is determined by 
one. two. or three pairs of heterozygous 
genes not showing dominance, there are (as 
in Figure 5-5 ) three, five, or seven possible 
phenotypic classes, respectively. As a result 
of dominance, however, the number of 
classes is reduced (cf. Chapter 4, p. 51). 
Since the estimated number of gene pairs 
responsible for a phenotype is directly re- 
lated to the number of phenotypic classes, 
the number of gene pairs involved in a quan- 
titative trait is underestimated whenever 
dominance occurs. This effect is important 
because many genes show complete or par- 
tial dominance. 

One can construct a hypothetical case in 
which two pairs of genes both showing domi- 
nance can give much the same phenotypic 
result as one pair with no dominance. Sup- 
pose gene A (as AA or Aa) adds 2 units 
of effect and its recessive allele a (as aa) 
adds only 1 unit; suppose B (as BB or Bb) 
subtracts 1 unit of effect and its recessive 
allele b (as bb) has no effect at all. Then 
a 2-unit individual {A A bb) mated with a 
0-unit one (aa BB) will give all intermediate 



1 -unit F| (AaBh). The F_. from the mat- 
ing of the F] can be derived by a branching 
track as shown in Figure 5-6. The pheno- 
typic ratio obtained in F 2 of 3:10:3 might 
be, in practice, difficult to distinguish from 
the 1:2:1 ratio obtained from crossing 
monohybrids showing no dominance. 3 

Dominance has a second effect with re- 
gard to quantitative traits; this can be illus- 
trated by means of two crosses involving the 
genes just described. In the first, two 0- 
unit individuals are crossed, aa Bb X aa Bb, 
yielding % aa B- (0 unit) and % aa bb 
(1 unit). In this case the parents, which 
are at one phenotypic extreme (0 unit), 
produce offspring which are, on the average, 
less extreme (0.25 unit). In the second 
case, two 2-unit individuals are crossed, 
Aa bb X Aa bb, yielding % A- bb (2 units) 
and % aa bb (1 unit). Here the parents 
are at the other phenotypic extreme (2 units) 
but produce offspring which are, on the 
average, less extreme (1.75 units). These 
results demonstrate regression, the conse- 
quence of dominance which causes individ- 
uals phenotypically extreme in either direc- 
tion to have progeny less extreme. 

Figure 5-7 illustrates the principle of re- 

3 See J. H. Edwards ( 1960). 



Multiple Alleles; Multigenic Traits 



65 



gression in polygenic situations. When no 
dominance occurs, the average offspring 
from parents at A, B, and C will be at the 
corresponding points A'. B\ C\ respectively, 
in the offspring curve. (The environment 
will cause some fluctuation around these 
phenotypic mean points in the offspring 
curve.) In the case of dominance, how- 
ever, the offspring of A will be, on the aver- 
age, to the right of A, as shown by arrows, 
whereas the offspring of C will generally be 
to the left of C. Contrary to what one might 
expect, the loss of extreme individuals gen- 
eration after generation will not make the 
entire population more and more homo- 
geneous phenotypically; there will be a 
closely counterbalancing tendency for the 
average members, B, of the population to 
produce offspring more extreme than them- 
selves in either direction. The result is that, 
as in cases of no dominance, the distribution 
curve for the offspring will be the same as 
for the parent population. 



PARENT 
GENERATION 



Group Selected 
as Parents 




Mean 



Mean of 
selected group 



OFFSPRING 
GENERATION 




FIGURE 5- 

character. 



Mean 



8. Selection for a quantitative 



PARENTS 




7^TS<7\ 



OFFSPRING 




figure 5-7. The principle of regression. 



To obtain a line of phenotypically extreme 
individuals from a population showing a 
quantitative trait, one would choose the ex- 
treme individuals as parents (Figure 5-8). 
If dominance were absent, the very first off- 
spring generation would have the same mean 
as the group selected as parents. Some de- 
gree of dominance usually occurs, and hence 
regression will usually occur; and the mean 
size of the first generation offspring will be 
somewhat less extreme than that of the se- 
lected parents, but somewhat more extreme 
than the original mean. As one continues 
to select appropriately extreme individuals 
as parents, the offspring in successive gen- 
erations will, on the average, approach more 
and more closely the extreme phenotype 
desired. 



66 CHAPTER 5 

SUMMARY AND CONCLUSIONS 

\ gene can cxisl in any one of two or more genetic states. These alternatives comprise 
a multiple allelic (sometimes isoallelic) series. 

Some alleles ol a given gene may produce apparently different qualitative phenotypic 
effects (and therefore show no dominance when hybrid); other alleles may produce 
different degrees of a quantitative phenotypic effect (in which case they may or may 
not show dominance when hybrid). 

Genes are the basis for both continuous and discontinuous traits. Continuous traits 
are usually determined by many gene pairs, each of which has a phenotypic effect that 
is small and often matched or exceeded by the action of the environment. 

The variability of a quantitative trait is such that the larger the number of heterozy- 
gous polygenes determining it, the narrower is the distribution curve and the smaller 
the chance of recovering either of the extreme phenotypes in the offspring. When 
polygenes are heterozygous, dominance has the effect of reducing the number of pheno- 
typic classes and of placing proportionally more offspring in extreme classes. Conse- 
quently, dominance usually causes one to underestimate the number of genes determin- 
ing a quantitative trait. Dominance also causes regression, so that selection must be 
continued for a number of generations to obtain a line which approaches the desired 
phenotype. 



REFERENCES 

Crow, J. F., Genetics Notes, 5th Ed., Minneapolis: Burgess, 1962. 

Edwards. J. H.. "The Simulation of Mendelism," Acta Genet., Basel, 10:63-70. 1960. 

Falconer, D. S.. Introduction to Quantitative Genetics, New York: Ronald Press, 1961. 

Race. R. R.. and Sanger. R.. Blood Groups in Man, 4th Ed., Philadelphia: F. A. Davis, 
1962. 

Wiener, A. S., and Wexler. I. B.. Heredity of the Blood Groups, New York: Grune 
& Stratton. 1958. 



QUESTIONS FOR DISCUSSION 

5.1. Discuss the occurrence of dominance with respect to blood group types. 

5.2. Why was it necessary to assume that a gene may have more than two allelic 
forms? 

5.3. A baby has blood type AB. What can you tell about the genotypes of its parents? 
What would you predict about the blood types of children it will later produce? 

5.4. If one parent is A blood type and the other is B, give their respective genotypes 
if they produced a large number of children whose blood types were: 

(a) All AB (c) Half AB, half A 

(b) Half AB, half B (d) h AB. Va A. *» B. U O 



Multiple Alleles; Multigenic Traits 67 

5.5. Give examples of complete dominance and of no dominance as found in human 
beings. 

5.6. Is the occurrence of complete dominance helpful in determining the genetic basis 
of alternatives for a given trait? Explain. 

5.7. A father with blood group types M and O has a child with MN and B blood types. 
What genotypes are possible for the mother? 

5.8. Criticize the statement: "Genes can be explained on the basis of a presence-or- 
absence hypothesis." 

5.9. A woman belonging to blood group B has a child with blood group O. Give their 
genotypes and those which, barring mutation, the father could not have. 

5.10. What do you think of the view that all the different genes that exist can be 
described as being different multiples of a single basic unit which is capable of 
retaining its integrity and is able to self-replicate? 

5.11. How many different genotypes are possible when there are four different alleles 
of a single gene? 

5.12. Does the discussion of multiple alleles in the text imply that: (a) There is an 
infinite variety of isoalleles? (b) No two genes are ever identical? Explain. 

5.13. Describe how you would test whether the genes for white eye color in two 
different populations of Drosophila were alleles, isoalleles, or nonalleles. 

5.14. In rabbits the following alleles produce a gradation effect from full pigmentation 
to white: agouti (C), chinchilla (c rh ) and albino (c). Another allele, c h , pro- 
duces the Himalayan coat-color pattern. C is completely dominant to all these 
alleles, c h is completely dominant to c, whereas c clt shows no dominance to 
c h or c. 

(a) How many different diploid genotypes are possible with the alleles men- 
tioned? 

(b) A light chinchilla mated to an agouti produced an albino in F x . Give the 
genotypes of parents and F 1 . 

(c) An agouti mated to a light chinchilla produced in F x one agouti and two 
Himalayan. Give the genotypes possible for parents and Fj. 

(d) An agouti rabbit crossed to a chinchilla rabbit produced an agouti offspring. 
What genotypic and phenotypic results would you expect from crossing the 
F 1 agouti with an albino? 

5.15. For each of the following matings involving Nicotiana give the percentage of 
aborted pollen tubes and the genotypes of the offspring. 

$ 9 6 9 

(a) sls2Xsls3 (c) si s4 X si s4 

(b) sls3Xs2s4 (d) s3 s4 X s2 s3 

5.16. Could you prove the existence of multiple allelism in an organism that only re- 
produces asexually? Explain. 

5.17. Do the genes for quantitative traits show epistasis? Explain. 

5.18. Does the environment have a more important role in determining the phenotype 
in cases of quantitative than in cases of qualitative traits? Explain. 

5.19. Under what circumstances are only seven phenotypes possible when three pairs 
of genes determine a quantitative trait? 

5.20. Discuss the statement: No new principles of genetics have originated from the 
study of polygenic traits. 



68 CHAPTER 5 

5.21. Suppose each gone represented by a capital letter causes a plant to grow an 
additional inch in height, aabbceddee plants being 12 inches tall. Assume 
independent segregation occurs tor all gene pairs in the following mating: 
A a 1111 cc /)r/ 1.1. <aa bb CC Dd Ee. 

(a) How tall are the parents? 

(b) HOW tall will the tallest F, be? 

(c) How tall will the shortest F, be? 

(d) What proportion of all F, will be the shortest? 

(e) Is dominance and/or epistasis involved in this system? Explain. 

5.22. In selecting for a quantitative trait, is the desired phenotype established in a 
pure line more easily when dominance does or does not occur? Explain. 

5.23. Measure the length of 10 lima beans to the nearest millimeter. Calculate the 
variance of this sample. To what can you attribute the variance? 

5.24. Is it of any advantage to an organism to have a trait determined quantitatively, 
that is. by many gene pairs, rather than qualitatively, that is, by principally one 
or a few gene pairs? Why? 

5.25. How would you prove that you were dealing with multiple alleles rather than 
multiple pairs of genes? 

5.26. In cattle a cross of a solid-coat breed and a spotted-coat breed produces a solid 
coat in Fj. Among the individuals of the spotted breed there is considerable 
variation, ranging from individuals that are solid-colored except for small white 
patches to those that are white with small colored patches. Selection within this 
breed can increase or decrease the colored areas. Discuss the genetic basis for 
coat color in these two breeds of cattle. 

5.27. Discuss the number of gene pairs involved in the following case: Golden Glow 
corn has 16 rows of kernels to the ear; Black Mexican has 8 rows. The F r is 
phenotypically intermediate, having an average of 12 rows. The F 2 is pheno- 
typically variable, ranging from 8 to 18 rows, with approximately one of each 
32 ears being as extreme as either P,. 

5.28. The Sebight Bantam and Golden Hamburgh are pure lines of fowl which differ 
in weight. Although the F, of crosses between these lines are fairly uniform and 
intermediate in weight, one in about every 150 F 2 is clearly heavier or lighter 
than either P 1 pure line. Suggest a genetic explanation for these results. 



Chapter *6 

PHENOTYPIC EFFECTS 
OF GENE ACTION 



Ti 



|he variations in phenotypic 
ratios that result from the in- 
teraction of alleles and non- 
alleles has been discussed (pp. 49-52). 
Several other phenotypic consequences of 
gene action are discussed in the present 
chapter. Let us begin with a discussion of 
viability effects, which also modify pheno- 
typic ratios. 

Viability Effects 

In the snapdragon (Antirrhinum) one finds 
two kinds of full grown plants, green and a 
paler green called auria. Green crossed by 
green produces only green, but auria by 
auria produces seedlings of which 25% are 
green (AA), 50% auria (Aa), and 25% 
white (aa). The last type of seedling dies 
after exhausting the food stored in the seed, 
because it lacks chlorophyll. Among full 
grown plants, the phenotypic ratio observed 
is y 3 green: % auria. In this case, the ab- 
sence of dominance gives the 1:2:1 ratio 
characteristic of a cross between monohy- 
brids in the seedling stage which, following 
the death of the albino, becomes a 2 : 1 ratio 
among the survivors. 

In mice, matings between yellow-haired 
individuals produce F : in the ratio 2 yel- 
low : 1 nonyellow. It is found after this mat- 
ing that % of m e fertilized eggs which 
should have completed development fail to 
do so and abort early in embryogenesis. 
Since crosses between nonyellows produce 
only nonyellows, the nonyellow phenotype 
69 



must be due to one type of homozygote, 
yellow must be a heterozygote, and the 
aborting individuals must be due to the other 
type of homozygote. The gene symbols usu- 
ally employed are not satisfactory here, for 
we now must describe two effects for each 
gene — color and viability. Moreover, the 
allele which is dominant for the first effect 
is recessive for the second, and vice versa. 
This problem is solved by using base letters 
with superscripts for each gene (Figure 
6-1), where the base letter refers to one 
trait and the superscript refers to the other 
trait. Let the superscript / be the recessive 
lethal effect of the gene dominant for yellow, 
Y, and the superscript L be the dominant 
normal viability effect of the allele recessive 
for nonyellow, y. Accordingly, the Fi from 
crossing two yellow mice (Y'y L X Y l y L ) are 
1 Y'Y 1 (dies) :2 Y'y L (yellow) : 1 y L y L (non- 
yellow). 

In both the snapdragon and mouse cases 
described, death results from the presence 
of a gene in homozygous condition. Genes 
that kill the individual before maturity are 
called lethal genes or lethals — those doing 
so only when homozygous are recessive 
lethals, and those acting in this way when 
heterozygous are dominant lethals. Lethals 
may act very early or very late in develop- 



yellow 
Y l y L 



'/2Y 1 , Vay 1 



'AY'Y 
dies 



ViY'y 1 - 
yellow 



yellow 



Y 1 y 



VjY 1 , Viy L 



v 4 y L y L 

nonyellow 



figure 6-1. Results of matings between yel- 
low mice. 



70 



CHAPTER 6 



SUPRA-VITAL 




SUB-VITAL 


SUB-LETHAL 


(beneficial) 




(detrimental) 


(semi-lethal) 


NORMAL 




i 

i 


LETHAL 


, VIABILITY 




1 


i i 


j 




A 


/\! 


V >! N 


1 1 i 1 
1.3 1.2 1.1 1.0 .9 


1 
.8 


1 1 1 1 1 1 II 

7 .6 .5 .4 .3 .2 .1 


RELATIVE 


VIABILITY 





FIGURE 6-2. Classification of c fleets that mu- 
tants have on viability. 

merit, or at any stage in between. Some- 
times a lethal effect is produced not by one 
gene or a pair, but by the combined effect 
of several nonallelic genes. In such a case, 
some of the nonalleles are contributed by 
each parent, and the offspring dies because 
the nonalleles, viable when separate, are 
lethal when together. 

Different alleles, recessive or dominant, 
have been shown to affect viability in dif- 
ferent degrees. These effects cover the en- 
tire spectrum — ranging from those which are 
lethal, through those which are greatly or 
slightly detrimental, to those which are ap- 
parently neutral or even beneficial (Figure 
6-2). When different combinations of 
alleles or nonalleles have different viabilities, 
the phenotypic ratios observed may differ 
significantly from those expected. The im- 
portance of the precautions to be taken, 
relative to the viability and fertility of the 
individuals bred in experiments designed to 
establish principles of transmission genetics, 
has already been discussed in Chapter 3. 

Pleiotropism 

Does each gene affect only one trait or can 
it have multiple, manifold, or pleiotropic 
effects? 

An investigation ' can be undertaken to 
answer this question, using two strains of 
Drosophila that are practically identical ge- 
netically (isogenic), except that one is pure 

1 See Th. Dobzhansky and A. M. Holtz ( 1943 ). 



tor the gene for dull-red eye color (vv+) 
and the other is pure for its allele white (w). 
Another trait is then chosen for examination 
in these two strains, one apparently uncon- 
nected with eye color. The trait selected is 
the shape of the spermatheca, an organ found 
in females which is used to store the sperm 
received. The ratio of the diameter to the 
height of this organ is determined for each 
of the two strains. This index of shape is 
found to be significantly different in the dull- 
red as compared to the white strain. From 
this result it can be concluded that the eye- 
color gene studied is pleiotropic. The re- 
sults of other studies also show many differ- 
ent genes to be pleiotropic for morpholog- 
ical traits. 

In Drosophila a recessive lethal gene 
called lethal-translucida causes pupae to be- 





# $ # # 



A- A 



fi # J? * ./ £ 



t r g & 






<c <^ ~ A" 4T * 



o 

figure 6-3. Pleiotropism at the biochemical 
level. (After E. H adorn.) 



Phenotypic Effects of Gene Action 



71 






figure 6-4. Silhouettes showing various types of human red blood cells: normal, in 
normal homozygote {A), sickle cell trait, in mutant heterozygote (B), sickle cell disease, 
in mutant homozygote (C). 



come translucent and die. Using suitable 
techniques, one can compare the kinds and 
amounts of chemical substances in the blood 
fluid of normal larvae and pupae with those 
found in the recessive lethal homozygotes 
(Figure 6-3). Some substances are found - 
in equal amounts in both genotypes (pep- 
tide III), others are more abundant in the 
lethal than in the normal individual (pep- 
tide I, peptide II, and proline), still others 
are less abundant (glutamine) or absent 
(cystine) in the lethal. Thus it is clear that 
pleiotropism also occurs at the biochemical 
level. 

In the case of the yellow mouse, the allele 
producing yellow coat color as a dominant 
effect also has a recessive lethal effect. On 
the presumption that homozygotes for this 
allele would have yellow body color had they 
survived, and on the basis that there is no 
obvious relation between coat color and via- 
bility, it could be concluded that this too is 
a case of pleiotropism. 

The coat color of Himalayan rabbits 
(cV') is usually mosaic; or black at the ex- 
tremities and white elsewhere (Chapter 1). 
Has this allele different effects on the same 
color trait in different parts of the coat? 
Because individuals with this genotype are 
completely black when grown under cold 

2 Based upon work of E. Hadorn. 



temperatures, we suspect that the gene has 
only one effect. This hypothesis is sup- 
ported by the finding that this genotype pro- 
duces an enzyme, necessary for pigment for- 
mation, which is temperature sensitive, being 
inactivated by temperatures above about 
34° C. Thus, in a cool climate, the body 
temperature is less than 34° C at the ex- 
tremities, and pigment is produced there; on 
the warm parts of the body no pigment is 
formed because the enzyme is inactivated by 
heat. The Himalayan pattern is attributed, 
then, to a single product of gene action 
which, because it is subject to modification 
by the environment, can result in two dif- 
ferent phenotypic alternatives of the same 
trait. 

A genetic disease in man called sickle-cell 
anemia is due to homozygosity for a certain 
allele. This disease involves the following 
effects, either singly or in any combination: 
anemia, enlarged spleen, skin lesions, heart, 
kidney, and brain damage. As a conse- 
quence, homozygotes for the gene for sick- 
ling usually die as adolescents or young 
adults; this allele, therefore, almost always 
acts as a recessive lethal. 

It is also found that the red blood cells 
of these homozygotes may become sickle- 
shaped instead of being disc-shaped (Figure 
6-4). Sickle-shaped cells can clump to- 
gether and clog blood vessels in various parts 



72 



CHAPTER 6 



of the body, leading to the malfunctions oi 

all the organs already mentioned; in addi- 
tion, these defective corpuscles are readilj 
destroyed by the body, with consequent 

anemia. 

We see. then, that the apparently unre- 
lated phenotypic effects of the gene for 
sickling are merely consequences of the sick- 
ling of red blood cells. Moreover, biochem- 
ical studies show that sickling itself is the 
result of the presence of an abnormal type 
of hemoglobin (having a slightly lower oxy- 
gen-carrying capacity than normal hemo- 
globin) which sickle-cell homozygotes carry 
in their red blood cells. There is. therefore, 
a pedigree of causes for the multiple effects 
of the gene for sickling. The first cause is 
the gene, the second is the abnormal hemo- 
globin it produces, the third is the sickling 
that follows, and the fourth is the subsequent 
red cell clumping and destruction which pro- 
duce gross organic defects and anemia. 

In this case all the pleiotropic effects are 
attributed to a single biochemical action of 
a gene. This single action then affects many 
varied chemical reactions involved in the 
production of different, at first apparently 
unrelated, traits. In view of the Himalayan 
rabbit and sickle-cell anemia studies, one 
may even hypothesize that most, if not all, 
genes have a single primary phenotypic ac- 
tivity. The pleiotropic effects described in 
other cases may yet prove, upon further 
analysis, to be tertiary or even further re- 
moved effects in a pedigree of causes, the 
primary cause being genie and the single 
secondary cause still undetermined. Reply- 
ing to the question with which this section 
started, the simplest hypothesis is that most, 
if not all, genes have only one phenotypic 
activity; this gene action has a pedigree of 
effects which results in pleiotropism. 

Penetrance and Expressivity 

Analysis of the genetic material has been 
greatly facilitated by the particular traits we 



have chosen to study. The most valuable 
kind of trait has been one based upon a 
genotype that always expresses itself in ap- 
proximately the same way, despite the nor- 
mal fluctuations of the environment. 

Consider, however, a pedigree for Poly- 
dactyly (Figure 6-5), a rare condition in 
which human beings have more than five 
digits on a limb. In the figure, the topmost 
female is affected, having five fingers on each 
hand and six toes on each foot. Her hus- 
band is normal in this respect. This couple 
has five children, three affected. This sug- 
gests that Polydactyly is due to a single domi- 
nant gene, P, and that the mother is Pp, the 
father pp. Consistent with this hypothesis 
is the result of the marriage of one of their 
affected daughters to a normal man. This 
marriage produced two sons, one of whom 
is affected, and this affected son, in turn, 
has five children including some affected and 
some unaffected. 

But now examine the left side of this pedi- 
gree. Note the firstborn son who is un- 
affected yet has an affected daughter. How 
may this be explained? It might be sup- 
posed that this son is genotypically pp and 
that his daughter is Pp, the P having been 
produced by mutation of p, then contributed 
to the daughter at conception. However, 
other pedigrees for Polydactyly also have 
cases in which two normal individuals have 
an affected child. Since Polydactyly is rare, 
mutations from p to P must be still more 
rare, so that the chance for such a mutant 
to appear in a sex cell of one of two normal 
parents is very small. It is most improbable, 
then, that such a rare mutation, if it occurs 
at random among normal individuals, would 
occur so often among the normals in pedi- 
grees for Polydactyly. 

A different explanation is that the firstborn 
son is in fact Pp, where P is not expressed 
in any detectable way, although it is ex- 
pressed in his daughter. This interpreta- 
tion is supported by the kind of expression 



Phenotypic Effects of Gene Action 



73 



a 



5.5 
6.6 



6~5^~i~~S~~±^3 



5.5 6.6 6.6 

6.6 5.5 5.5 



* 6 



5S 



5.6 
6.7 



i555i 



figure 6-5. A pedigree of Polydactyly in man. 



6.6 
6.6 



that the P gene produces in different affected 
individuals in this pedigree. These may 
have the normal number of fingers but have 
extra toes, or they may have the reverse; 
they may have different numbers of toes on 
the two feet, or they may have extra fingers 
on one hand and the normal number on the 
other. The expression of Polydactyly, so far 
as the number of extra digits is concerned, is 
clearly quite variable. Accordingly, since it 
is possible to have no expression on one 
limb of an individual known to be Pp it must 
also occur that, on occasion, expression fails 
on all four limbs of an individual with this 
genotype. 

The ability of a given gene or gene com- 
bination to be expressed phenotypically in 
one way or another is called penetrance. 
The P gene in heterozygous condition, there- 
fore, has a penetrance of less than 100%, 
sometimes failing to produce any detectable 
phenotypic effect when present. Although 
a polydactylous person is certain to carry P, 
a normal phenotype can represent either the 
Pp or pp genotype. Since Polydactyly is 
rare it is usually quite safe to score as pp 
the genotype of a normal individual who 
marries into a line of descent containing P. 



The expression of P when heterozygous is 
not only quite variable with respect to the 
number and position of extra digits, but 
further variability of expression is demon- 
strated by the different degrees of develop- 
ment which the extra digits show. The term 
expressivity is used to refer to the kind or 
degree of effect produced by a penetrant 
genotype. In individuals where P is non- 
penetrant when heterozygous there is no ex- 
pressivity, and when P is penetrant its ex- 
pressivity is variable. 

What factors are involved in the produc- 
tion of variable penetrance, or, in cases of 
penetrance, of variable expressivity? A 
study of a genetically uniform line of guinea 
pigs shows that Polydactyly occurs more fre- 
quently in the litters from younger than from 
older mothers. In this case the physiological 
changes accompanying age modify pene- 
trance. In another case, a genetically uni- 
form line of Drosophila flies shows a greater 
percentage of penetrance of an abnormal 
abdomen phenotype when moisture content 
during development is high than when it is 
low. Both of these examples illustrate that 
variations in penetrance can be produced by 
variations in the environment of different 



74 



CHAPTER 6 



individuals with essentially identical geno- 
types. 

We are already familiar with the effect ol 
genotypic variations upon penetrance under 
essentially constant environmental condi- 
tions. The penetrance of an allele may de- 
pend upon the nature of its partner allele 
in cases of complete or partial dominance, 
and the penetrance of one or a pair of alleles 
may be modified by its epistatic- hypostatic 
relations to nonallelic genes (Chapter 4). 
Similarly, it can be shown that variable ex- 
pressivity may be the consequence of dif- 
ferences in either or both the environment 
and the genotype. 

One should be careful to differentiate be- 
tween penetrance and expressivity on one 
hand and dominance and epistasis on the 
other. Suppose, with respect to height, TT 
always produces tall, TT always short, and 
TT always medium. Although the hybrid 
shows incomplete, partial, or no dominance, 
there is 100% penetrance for each of the 
three genotypes. If among the mediums 
there was some variability in size (due to 
variations in environment or the rest of the 
genotype), we would have different expres- 
sivities for the 100% penetrant TT geno- 
type. If, however, for the same reasons. 
TT sometimes produced a short individual. 
this would be a case of nonpenetrance of 
T in TT. 

The terms penetrance and expressivity 
were used to compare the phenotypic events 
that occur in different individuals genetically 
identical in one particular respect. That is, 
once any phenotypic expression occurred 
within an individual, the genotype was said 
to be penetrant, and all other phenotypic 
comparisons between penetrant individuals 
were considered matters of expressivity. In 
fact, however, one can also correctly speak 
about penetrance within an individual for 
those cases in which the particular genotype 
has two or more occasions to express itself. 
Thus, for example, the gene for Polydactyly 



has two apparently equal chances to be pene- 
trant in the case o\ the hands, and two ap- 
parently equal chances to be penetrant in 
the case o\ the feet. The genotype may be 
penetrant in one hand (six fingers) and not 
in the other (five lingers), it may be pene- 
trant in the feet (each foot having six toes 
— represented as 6.6) and not in the hands 
(5.5). When differences in penetrance (or 
expressivity) are shown by essentially dupli- 
cate parts of the same individual (one hand 
having seven and the other six digits, or one 
hand having a large and the other a small 
extra digit), one can be reasonably certain 
that these differences have an environmental 
and not a mutational basis. However, when 
different individuals are compared with re- 
spect to penetrance or expressivity, it is often 
impossible to attribute, with assurance, simi- 
larities or differences among them to geno- 
type or to environment, if both of these 
factors can vary in uncontrolled ways (as 
already implied on p. 12). 

Studies of Human Twins 

In organisms other than man experimental 
conditions can be controlled so that a stand- 
ard genotype exposed to different environ- 
ments shows to what extent environment is 
responsible for phenotypic variability, where- 
as a standard environment to which different 
genotypes are exposed reveals to what ex- 
tent these genotypes produce different 
phenotypes. Since neither the environment 
nor the genotypes of human beings are sub- 
ject to experimental control, how can we 
determine to what extent a particular human 
trait is controlled by genotype (nature) and 
by environment (nurture)? Fortunately, 
this nature— nurture problem can be studied 
using the results of certain naturally occur- 
ring phenomena. 

An individual contains many different 
parts which presumably have the identical 
genotype. Accordingly, as mentioned, one 
can attribute to nurture any phenotypic dif- 



Phenol y pic Effects of Gene Action 



75 



ferences in expressivity or penetrance found 
among parts that are essentially duplicates 
of each other. For example, a heterozy- 
gote for Polydactyly with six fingers on one 
hand and five on the other illustrates the ex- 
tent to which environment can affect this 
trait. When, however, a trait involves the 
entire body, or only one or several different 
nonduplicate parts of the body, the contri- 
bution of nurture can be learned only by 
comparing different individuals which have 
identical genotypes. 

Since each human individual is hetero- 
zygous for a relatively large number of genes, 
the chance of obtaining genetic identity in 
two siblings (children of the same parents) 
is very small indeed. However, two or more 
siblings with identical genotypes can be pro- 
duced in man by asexual reproduction, 
which occurs in the following manner. A 
single fertilized egg starts development nor- 
mally by undergoing a series of mitotic divi- 
sions. At some time, however, the cells 
produced fail to adhere to each other, as 
they would normally do, and separate into 
two or more groups, each of which may be 
able to develop into a complete individual. 
Each individual thus produced is, barring 
mutation, genetically identical to all others 
formed from the same fertilized egg. The 
separation referred to may occur at different 
stages of early development, and the num- 
ber of cells in the two or more groups formed 
may be unequal. Separation may even 
occur more than once, at different times in 
the development of a particular zygote. The 
individuals produced in this asexual manner 
are identical or monozygotic twins, triplets, 
quadruplets, etc. We need consider only 
identical twins here, since multiple births 
of greater number are usually too infre- 
quent to be useful for a general study of 
the nature-nurture problem. 

Multiple births can also be produced di- 
rectly by sexual reproduction. When twins 
are produced in this way, they start as two 



separate eggs, each fertilized by a separate 
sperm. Such twins are genetically different 
— being, in this respect, no more similar 
than siblings conceived at different times — 
and arc nonidentical or dizygotic (fraternal) 
twins. 

These two kinds of twins provide natural 
experimental material for determining the 
relative influence of genotype and environ- 
ment upon the phenotype. Barring muta- 
tion, monozygotic twins furnish the identical 
genotype in two individuals, and both kinds 
of twins share similar environments before 
birth and, when raised together, after birth. 

The phenotypic differences between iden- 
tical twins reared together are essentially the 
consequence of environment alone (Figure 
6-6). One can compare the average dif- 
ference between such identical twins with 
the average difference between identical 
twins who, for one reason or another, were 
reared apart. This comparison yields in- 
formation regarding the influence of greater, 
as compared with lesser, environmental dif- 
ferences upon the phenotype. Since non- 
identical or identical twins reared together 
are exposed to environments which, on the 
average, vary to the same extent, a com- 
parison of the average difference between 
identical twins and the average difference 
between nonidentical twins will give an index 
of the genotype's role in causing the differ- 
ences observed. In order to collect valid 
data from twin studies, it is essential that 
one be able to recognize in each case whether 
the twins are monozygotic or dizygotic in 
origin. 

The best way to identify nonidentical 
twins is to compare the siblings with refer- 
ence to a large number of traits known to 
have a basis in those genes which are 100% 
penetrant and of fairly uniform expressivity 
—such traits as sex, eye color, ABO, MN, 
Rh. and other blood group types. Naturally, 
only traits for which at least one parent is 
heterozygous are of use in testing the dizy- 



76 



CHAPTER 6 



gotic origin of twins. Assuming the absence 
oi mutation. an\ single difference in such 
traits would prove the twins aonidentical. 
(On this basis, twins oi' opposite sex are 
classified immediately as aonidentical.) Of 
course, two such differences would make the 
decision practically infallible, since two mu- 
tations in genes governing the limited num- 
ber of traits being compared in a pair of 
identical twins would be so rare as to be 
beyond any reasonable probability of occur- 
rence. 

Identification of monozygotic twins re- 
quires the same phenotypic comparisons. 
The larger the number of traits for which 
no genetic difference is demonstrated, the 
greater the probability that the twins are 
identical. When the number of traits serv- 
ing to test the genotypes of twins is suffi- 
ciently large, it becomes nearly certain that 
they would have shown one or more differ- 
ences had they been dizygotic in origin. 
Failure to show any such difference, then, 
may be attributed to identical genotypes de- 
rived from a single zygote. 

Let us outline the procedure one might 
actually follow in using twins to study the 
relative roles of genotype and environment 
in producing specific traits. The objective 
is to score separately the percentage of iden- 
tical and nonidentical twin pairs reared to- 
gether in which one or both siblings have 
the trait under consideration. Suppose one 
wished to study the ABO blood group in 
this respect. One would determine the per- 
centage of concordance, that is, the percent- 
ages of identical and of nonidentical pairs 
in which both members of a pair have the 
same phenotype. In the case of identical 
twins concordance for ABO blood type is 
found to be 100%. 

In determining concordance for noniden- 
tical twins one usually scores only pairs in 
which the twins are of the same sex. This 
convention is necessary because the post- 
natal environment of twins of opposite sex 




FIGURE 6-6. Identical twins, Ira and Joel, at 
3% months, at 8 years, and at 19 years of age. 
(Courtesy of Mrs. Reida Postrel Herskowitz 
July 14, 1946.) 



Phenotypic Effects of Gene Action 



ABO BLOOD GROUP 

CLUBFOOT 

TUBERCULOSIS 

PARALYTIC POLIOMYELITIS 



IDENTICAL 



77 
NON-IDENTICAL 



100 


32 


74 


36 



64 



28 



figure 6-7. Discordance (unshaded) and percentage con- 
cordance (shaded) for various physical traits in twins reared 
together. 



is likely to be more different than that of 
twins of the same sex. (If the environment 
differed for the two kinds of twins, one 
would not be able to specify whether the 
environment or the genotype was the cause 
of a phenotypic difference that is greater 
among nonidenticals than identicals.) Only 
twins of the same sex are used in the twin 
studies discussed here. 

The concordance for ABO blood type is 
approximately 64% for nonidenticals. Had 
concordance been the same (64% or 
100%) for both types of twins, we would 
conclude that there is no net genetic or 
environmental difference for ABO blood 
group in the two types of twins. The con- 
cordances observed do differ, however, and 
do so in a particular direction. Because of 
this difference the 100% concordance for 
identicals must mean that this trait is deter- 
mined genetically with a penetrance of 
100% despite the environmental fluctua- 
tions normally occurring between identical 
twins. Since an equivalent amount of en- 
vironmental fluctuation caused no differences 
in the case of identicals, the lower percent- 
age of concordance for nonidenticals cannot 
be attributed in any part to environment. 
This lower concordance must be attributed, 
therefore, to the differences in genotype 
which nonidenticals can have in this respect. 



Of course, we could have predicted such a 
result from the previous knowledge that 
ABO blood type is genetically determined 
and is known to have complete penetrance. 
The lower concordance for nonidenticals, 
therefore, must be due to their receiving 
different genotypes from parents, one or 
both of whom were heterozygous for l A 
or P. 

It is theoretically possible to obtain a re- 
sult in which concordance is lower for iden- 
ticals than it is for nonidenticals. Such a 
difference in concordance could be ascribed 
to environmental differences being greater 
among the identicals than among the non- 
identicals. 

Consider the results of concordance stud- 
ies for some physical traits in twins (Figure 
6-7). Concordance for clubfoot is 32% 
for identicals, but only 3% for nonidenticals. 
The extra concordance of 29% (32% minus 
3% ) found among identicals must be attrib- 
uted to their identical genotype. The 3% 
concordance found among nonidenticals 
might be due entirely to similarity in geno- 
type or entirely to the environment, or to 
some combination of these two factors. 
Since we cannot decide from these data, we 
conclude that in twins or other individuals 
exposed to the same environment that twins 
are, the occurrence of clubfoot can be at- 



78 



CHAPTER 6 



tributed to genotype approximately 299? of 
the time, with 329? as the approximate 
upper limit. 

In the case of the identicals. 689? of the 
time the second twin tailed to have clubfoot 
when the first twin did. The failure of con- 
cordance is called discordance. The 68% 
discordance between identicals is attributable 
to differences in their environment. It is 
concluded, then, that in twins or other indi- 
viduals exposed to the same environment 
that twins are. the occurrence o\ clubfoot is 
the result of the environment approximately 
68% of the time, with 7192 as the approxi- 
mate upper limit. 

Concordance-discordance studies reveal 
only the relative contributions of genotype 
and environment to a particular phenotype 
(clubfoot, for example, as in the case just 
discussed). Such studies do not teach us 
anything about the kinds of environment 
involved when the genotype determines the 
phenotype under consideration, nor do they 
teach us anything about the genotypes in- 
volved when the environment decides the 
phenotype. The clubfoot twin studies also 
tell us nothing about the effect upon pene- 
trance of clubfoot caused by environmental 
differences greater than those found between 
twins reared together. Application of the 
conclusions from twin studies to the gen- 
eral population assumes that environments 
for twins and nontwins are the same. Such 
an assumption may be invalid. 

In the case of tuberculosis, concordance 
is 74% for identicals and 28 % for non- 
identicals. Accepting the supposition that 
both types of twins have the same average 
exposure to the tubercle bacillus, the sus- 
ceptibility to this disease is determined ge- 
netically 46 to 74% of the time and environ- 
mentally 26 to 54% of the time. In sup- 
port of the view that the extra concordance 
among identicals has a genetic basis is the 
finding that concordant identicals usually 



have the same form of this disease, affecting 
corresponding organs with the same severity, 
whereas this similarity is less frequent among 
concordant non identicals. 

In earlier studies, paralytic poliomyelitis 
was 36% concordant for identicals and 6% 
concordant for nonidenticals. As in the 
case of tuberculosis, the occurrence of the 
disease probably did not depend upon the 
infective organisms because most human be- 
ings were exposed to them normally. Ac- 
cordingly, the incidence of this disease de- 
pended upon the rest of the environment 
64 to 70% of the time and the genotype 
30 to 36% of the time. In the case of 
measles, the fact that concordance is very 
high among both types of twins simply means 
that any genetic basis for susceptibility to 
this disease is quite uniform throughout the 
population from which the twin samples 
were obtained. 

The relative contributions of genotype and 
environment to personality and other mental 
traits can also be studied by the twin method. 
When a metronome is run at a series of dif- 
ferent speeds, the tempo preferred by differ- 
ent persons is different. Tempo preference 
may be considered to be one aspect of the 
general personality. When tests are made 
to compare the tempo preferred by identical 
twins, the difference in their scores is found 
to be 7.8 of the units employed (Figure 
6-8). This is, as might be expected, not 
significantly different from the difference in 
score of 8.7 units obtained by testing a 
given individual on different occasions. 
However, nonidenticals have a difference in 
score of 15, which is significantly different, 
being about twice that of the identicals. 
Since nontwin siblings have a difference in 
score of 14.5, they prove to be as similar in 
this respect as are nonidentical twins. 
Finally, unrelated persons show a difference 
in score of 19.5 units. Since the greater 
the genetic similarity the smaller the differ- 



Phenotypic Effects of Gene Action 



79 



INDIVIDUALS 


DIFFERENCE 
IN SCORE 


Same person on 
different occasions 


8.7 


Monozygotic twins 


7.8 


Dizygotic twins 


15.0 


Siblings 


14.5 


Unrelated 


19.5 



figure 6-8. Variation in preferred tempo. 
(After C. Stem.) 



ence in score, there is clearly a genotypic 
contribution to this personality trait. 

Studies of twins for the mental disease 
schizophrenia show concordance of 86% for 
identicals and 14% for nonidenticals. How- 
ever, it is likely that the environment is not 
the same for both types of twins, differences 
in social environment causing more discord- 
ance in the case of nonidenticals than in the 
case of identicals. Nevertheless, in support 
of the view that not all the concordance for 
identicals is attributable to their similar en- 
vironment and that some genotypic basis 
exists for concordance are two cases of iden- 
tical twins who were separated, raised in dif- 
ferent environments, yet were concordant at 
about the same age. 

Different people, of course, score differ- 
ently on I.Q. examinations. The differences 
in ability to answer questions on these exam- 
inations can be used to measure what may 
be called test intelligence. Although the 
scores of nonsiblings vary widely above and 
below 100, the difference between the scores 
of twins reared together is only 3.1 for iden- 
ticals but 7.5 for nonidenticals. Clearly, 
identity in genotype makes for greater simi- 
larity in score. Identicals reared apart have 
scores that differ by 6. In this case the 



greater difference in environment makes for 
a greater difference in performance of iden- 
ticals, but this is still not so great a difference 
as is obtained between nonidenticals reared 
together. Therefore, both genotypic and en- 
vironmental factors affect the trait, test in- 
telligence. 

In the case of ABO blood group we have 
already discussed the nature of the genetic 
factors involved. Although the twin and 
other methods used in this section tell 
whether genotypic differences are associated 
with the occurrence of the other phenotypes 
considered, they provide no information re- 
garding the location, number, or recombina- 
tional properties of the genes involved. 

Developmental Effects 

Many of the mutants present at fertiliza- 
tion in multicellular plant and animal forms 
are detected by some visible change they 
produce in morphology. This phenotypic 
change is usually macroscopic and is first 
noted a considerable time after the organism 
starts its development. How does the mu- 
tant change normal development in order to 
produce the new morphological result? The 
answer to this question involves the manner 
in which phenotypes (of any type) come 
into being via gene action, and is the subject 
of phenogenetics, one aspect of which is 
developmental genetics. 

Consider the genetic and phenogenetic in- 
formation obtained from studying one par- 
ticular case. In the chicken a novel type 
occurs whose legs are so short that they give 
the impression that the bird is creeping. 
This abnormal "Creeper" phenotype and 
the normal phenotype 3 can be seen in the 
roosters in Figure 6-9. 

The genetic study of this phenotype gives 
the following results: reciprocal crosses of 
Creeper by normal produce a 1 : 1 ratio of 

3 Studied by W. Landauer. V. Hamburger, D. Rud- 
nick, and L. C. Dunn. 



so 



CHAPTER 6 




FIGURE 6-9. Normal (right) and Creeper (left) roosters. (Courtesy of L. C. 
Dunn; reprinted by permission of McGraw-Hill Book Co., Inc., from Study 
Guide and Workbook for Genetics by I. H. Herskowitz. Copyright, 1960.) 



Creeper: normal chicks; Creepers crossed 
with Creepers give, in the adult stage, 
775:388 as Creeper: normal, a result which 
is considered an excellent fit to a 2: 1 ratio. 
It is reasonable to suppose, therefore, that 
Creeper is heterozygous for a single pair of 
segregating genes, in which the Creeper gene, 
Cp, is dominant to its normal allele, +. 
The 2 : 1 ratio is taken to indicate that the 
mutant homozygote, Cp Cp, is lethal. The 
possibility that Cp acts as a recessive lethal 
is supported by a comparison of the survival 
frequency of embryos having normal parents 
with that of embryos having parents both of 
which are Creeper. It is found that about 
25% more embryos die within three days 
of incubation in the latter than in the former 
case. 

What is the developmental, phenogenetic 
basis for Cp Cp, which acts as a recessive 
lethal; Cp +, which produces Creeper; and 
-| — |-, which produces normal? Although 
Cp Cp usually dies within three days of in- 
cubation, on rare occasions it survives 19 
days, about the time of hatching from the 
shell. Such a rare Creeper homozygote is 
shown at the left of Figure 6-10 (the com- 
parable normal individual is at the right) 



and possesses the following syndrome of 
malformations: the eyes are split, smaller 
than normal, and have no eyelids; the head 
is misshapen, and the body is smaller; the 
skeleton is not ossified and — as seen on top 
of the black paper used as background in 
the figure — only the digits of the limbs are 
well formed. 

A study of Cp -j- development shows that, 
at seven days of incubation, the leg buds 
are shorter than in normal embryos. This 
morphological manifestation of Cp action 
must be based upon events occurring earlier 
in development, for at 48 hours of incuba- 
tion (Figure 6-11), a Cp -\- embryo (left) 
is smaller, less developed, and does not have 
the head flexure already present in a H — (- 
embryo (right). In fact, differences like 
this can be seen even twelve hours earlier, 
i.e., at 36 hours of incubation. 

In both the homozygote and heterozygote 
for Cp, the differentiation of cartilage has 
been disturbed. The Cp -+- individual has 
the disease called chondrodystrophy (or 
achondroplasia) and the CpCp individual 
has the cartilage disease, phocomelia (see 
p. 4). Both diseases were recognized in 
human families more than a hundred years 



Phenotypic Effects of Gene Action 



81 




figure 6-10. Normal (right) and Creeper (left) homozygote at about 19 
days of development. (Courtesy of L. C. Dunn; reprinted by permission of 
McGraw-Hill Book Co., Inc., from Study Guide and Workbook for Genetics 
by I. H. Herskowitz. Copyright, 1960.) 



figure 6-11. Normal (right) and Creeper (left) heterozygote embryos at 
about 48 hours of development. (Courtesy of L. C. Dunn; reprinted by per- 
mission of McGraw-Hill Book Co., Inc., from Study Guide and Workbook 
for Genetics by I. H. Herskowitz. Copyright, I960.) 




S2 



CHAPTER 6 



ago; when both parents arc chondrodys- 

trophic, some children are phocomelic and 

have severelj deformed limbs. The condi- 
tion observed in such individuals can be at- 

tributed to the presence of a mutant gene 
(like the Cp gene in fowl) in a double dose, 
that is. when homozygous. 

It was already mentioned that at 36 hours 
of incubation. Cp - individuals develop 
more slowlj than -f + individuals. In nor- 
mals at this stage the tissue for the hind 
limb buds grows very rapidly, whereas other 
tissues grow more slowly. If some of the 
effect of Cp in single or double dose is to 
cause a generalized slowing-down of growth, 
the structures most affected will be those 
growing most rapidly at the time. Such a 
genetically-induced slowdown in growth rate, 
starting at about this particular time in de- 
velopment, is expected, therefore, to reduce 
the size of the hind limbs and the long bones 
of fore limbs. 

It should not be concluded, however, that 
the tissue for hind limb is completely passive 
to Cp action and that its sole response is the 
slowdown in growth rate. We can study 
the developmental fate of prospective hind- 
limb tissue by transplantation experiments. 
If such tissue from a normal chick embryo 
is transplanted to a more forward position 
in another normal chick embryo, it grows 
out as a normal limb. If, however, the pro- 
spective hind-limb tissue from a homozygous 
Creeper embryo is transplanted to a more 
forward position in a normal chick embryo, 
it grows out as a Creeper type leg. This 
result demonstrates that even at a very early 
stage, before there is any actual hind limb, 
prospective limb tissue from Creeper is 
already permanently determined by the 
Creeper genotype to develop as Creeper 
limb, that is. its competence to develop nor- 
mally is already lost. 

It also should not be assumed that all ab- 
normal tissues found in homozygous Creep- 
ers have been determined at an early stage 



in development and, thus, possess only the 
Creeper alternative. As mentioned, CpCp 
individuals have small, split eyes. The earl) 
eye anlage [imaginal disc) from a normal 
embryo can be transplanted to an abnormal 
position in a normal embryo. In this posi- 
tion it grows into an eye just like that of 
homozygous Creeper, but an eye anlage from 
a Cp Cp embryo, transplanted to the eye- 
forming region of a normal embryo, grows 
into a normal eye. We can conclude, there- 
fore, that the abnormal Creeper eye is due, 
not to some change in the competence of 
the eye tissue, but rather to some kind of 
abnormality in its surroundings. It can, 
therefore, be supposed that in the Creeper 
homozygote the normally competent eye 
anlage probably undergoes a kind of starva- 
tion due to the poor circulation that the 
genotype produces. Such a hypothesis is 
supported by two lines of evidence: first, 
most prospective tissues of Cp Cp placed on 
a complete culture medium /'// vitro grow 
quite normally, although heart tissue does 
not grow as well as normal heart tissue; 
second, when limb rudiments from normal 
embryos are grown in vitro in a nutritionally 
dilute culture medium, they develop many 
of the characteristics of the Cp Cp limbs. 

The study of Creeper fowl demonstrates 
that the multiple effects of this mutant which 
have been found at the completion of de- 
velopment are due to gene-directed changes 
originating much earlier in development. 
In fact, we can infer from the developmental 
fate of prospective limbs in Creeper embryos 
that a genotype produces changes that pre- 
cede morphological changes. The Creeper 
gene apparently modifies the physiology of 
the individual in such a way that general 
growth is slowed down, and the prospective 
fate of certain tissues is fixed, so that the 
morphological changes later noted are a di- 
rect consequence of these changes. The 
gene-caused physiological changes can be 
attributed, in turn, to changes in the bio- 



Phenotypic Effects of Gene Action 



83 



chemical reactions involved in cellular me- 
tabolism. 

The Creeper case apparently involves ge- 
netically determined metabolic changes that 
take place within certain cells (to produce 
an abnormal nutritional environment) and 
affect the functioning of other cells (the 
differentiation of eye tissue). Let us con- 
sider two groups of studies with mice to 
learn more about the genetic control of 
effects produced external to the cell in which 
the gene acts. One group of investigations 4 
involves a comparative study of normal and 
dwarf mice. The dwarfs have all of their 
body parts proportionally reduced in size, 
because of an apparently completely reces- 
sive gene in homozygous condition. During 
early development, both dwarf and normal 
mice grow at equal speeds, but later, the 
dwarf suddenly stops growing and never 
reaches sexual maturity. A microscopic 
study of the anterior pituitary gland shows 
that it is considerably smaller in the dwarf 
than in the normal mouse. Moreover, cer- 
tain large cells, normally present, are absent 
in dwarf pituitaries; apparently these are the 
cells that secrete growth hormone. That this 
is a case of genetically produced pituitary 
dwarfism is supported by the following type 
of experiment: using pairs of dwarf litter 
mates about 30 days old, one mouse of a 
pair is injected with extracts of pituitary 
glands from dwarf mice (Figure 6-12, B) 
each day for 30 days, whereas the other 
mouse is injected in a comparable way with 
extracts of pituitary glands from normal 
mice (Figure 6-12. A). During this period 
of treatment, the former mouse remains es- 
sentially dwarf, while the latter grows until 
it is virtually normal. Here, then, we are 
dealing with a chemical messenger, pituitary 
hormone, which regulates growth in general, 
and whose presence is dependent upon a 
single pair of genes. 

4 Based upon work of G. D. Snell, of P. E. Smith 
and E. C. MacDowell. and of T. Francis. 




44 49 54 

AGE IN DAYS 



figure 6-12. Effect of injecting pituitary 
viand extracts into dwarf mice. (See text for 
explanation. ) 



The second group of studies is concerned 
with mouse tails. The normal (+ +) 
mouse has a long tail; a particular mutant 
strain has a shortened tail (Brachyury, or 
Brachy). 5 Brachy crossed with Brachy pro- 
duces % Brachy: V 3 normal offspring, a ratio 
which suggests that the gene for Brachy, T, 
is dominant for short tails and recessive for 
lethality. Brachy mice should, therefore, be 
T -f. When the embryology of offspring 
produced from a mating between Brachys 
(T + X T +) is studied, about 25% of the 
embryos are normal (+ +), about 50% 
show tail degeneration by 1 1 days of de- 
velopment (7-f-), and about 25% of the 
embryos (TT) are monsters (Figure 6-13) 
which have misdirected posterior limb buds, 
zigzag neural tubes, and no notochord. 
Since their whole posterior part is undevel- 
oped, T T individuals cannot make a placen- 
tal connection and die after about 10 or 11 
days of development. 

Consider further the T T individual, whose 
segments, or somites, in the posterior part 

5 Based upon work of L. C. Dunn, P. Chesley, 
and D. Bennett. 



St 



CHAPTER 6 



of the bodj arc so grossly abnormal. Other 
embryological work shows that proper so- 
mite formation depends on the presence of 
presumptive notochorda! tissue. When nor- 
mal, presumptive notochord tissue is present, 
its surrounding normal mesoderm is induced 
to form cartilage and vertebral segments. 
It seems reasonable to attribute the failure 
of cartilage and vertebrae formation in T T 
individuals to the failure o( its presumptive 
notochord (which has lost the ability to de- 
velop into notochord) to induce the differen- 



tiation of mesoderm. This explanation is 
subject to test in certain experiments em- 
ploying tissue culture. Using tissues from 
normal mice, presumptive notochorda! tis- 
sue, which has mesodermal tissue from the 
same or another individual wrapped around 
it. is placed in the tissue culture medium. 
Development of cartilage and vertebral seg- 
ments occurs under these conditions. More- 
oxer, the mesoderm from normal embryos 
also develops into cartilage and vertebral 
segments, when surrounding presumptive 




BRACHY T+ 



25% 




die 10% days 




BRACHY T+ 



50% 




25% 



birth 



days 



MONSTER TT 



BRACHY T+ 



NORMAL + + 
(Size Reduced) 



figure 6-13. Brachyury in the house mouse. (Courtesy of L. C. 
Dunn; reprinted by permission of McGraw-Hill Book Co., Inc., from 
Study Guide and Workbook for Genetics by I. H. Herskowitz. Copy- 
right, 1 960.) 



Phenotypic Effects of Gene Action 



85 



notochord from young T T embryos; under 
similar conditions, however, mesoderm from 
T T embryos does not form cartilage or 
vertebrae when it surrounds presumptive 
notochord from normal embryos. We must 
conclude — contrary to expectation — that the 
normal inductive relationship is disturbed in 
T T because its mesoderm is incompetent to 
respond to the normal inductive stimuli of 
presumptive notochord tissue. 

The preceding discussion shows that ge- 
netic changes can influence or control de- 
velopment in multicellular organisms by 
modifying ( 1 ) the relative growth rates of 
parts (as in Creeper) and (2) the over-all 
growth rate (as in pituitary dwarfism) with- 
out affecting the competence of some or all 
of the tissues affected. Genetic changes can 



also affect differentiation by changing tissue 
competence (homozygous Creeper limbs). 
When adjacent tissues interact by induction, 
differentiation can be modified by gene- 
caused changes in competence to respond to 
inducing agents (nonresponsiveness of T T 
mesoderm to presumptive notochord) and, 
presumably, also by changes in inducing 
ability. It should be realized, however, that 
although differentiation and development in 
higher multicellular organisms involve inter- 
cellular interactions of all the types men- 
tioned, some cellular traits are produced 
solely through the intracellular action of the 
genotype. Such behavior occurs, for exam- 
ple, in mutants induced during embryogeny 
which have detrimental or lethal effects in 
the cells containing them. 



SUMMARY AND CONCLUSIONS 

Different alleles may produce detectable effects upon viability at any stage in the life 
history of an individual and may modify the expected phenotypic ratio so that certain 
classes of offspring are in excess, in reduced frequency, or are absent. The last effect 
is produced by dominant as well as recessive lethal genes. 

A gene usually produces effects upon a wide variety of morphological and biochemical 
traits. These pleiotropic effects are the consequence of a pedigree of causes traceable, 
in some cases, to a single action on the part of the gene. It is hypothesized that most, 
if not all, genes have only one, probably biochemical, phenotypic action. 

Penetrance and expressivity depend upon both the genotype and the environment. 
The traits most useful for the study of transmission genetics are those whose penetrance 
is 100% and whose expressivity is uniform when subjected to the normal variations 
of environment. In human beings the occurrence of essentially duplicate parts within 
an individual, and of identical and nonidentical twins, offers the opportunity to test 
the effect of environment and of genotype upon the penetrance and expressivity of a 
given phenotypic alternative. 

It has been shown that a considerable number of physical and mental traits are 
determined by the joint action of genotype and environment, sometimes one and some- 
times the other having the greater influence. 

The twin methods described do not study the transmissive characteristics of the 
genotypes involved. They do not, therefore, reveal anything regarding the location, 
number, or recombinational properties of genes. 

Phenogenetics, the study of how genetically determined phenotypes come into being. 



86 ( H \r 1 1 k 6 

can be investigated bj using morphological traits. Although phenogenetics often starts 
out as a study of the developmental genetics of morphology, the final morphological 
outcome which usualh is a pleiotropic one — is often found to be based upon earlier 
morphological changes which arc. in turn, preceded by still earlier-occurring physio- 
logical changes. Consequently, the developmental genetics ol morphological features 
is based upon gene-caused physiological changes and leads to a study of physiological 
genetics. 

Physiological genetics reveals that the physiological effect o\ the genotype is some- 
times intracellular and sometimes intercellular. The gene-based action that certain 
cells have on others can involve a general or localized control of growth rates and dif- 
ferentiation. This action can occur nearby, via induction; or at a distance, by means 
of a general nutritive effect, by hormones, and probably by nerve impulses and muscular 
contractions. Gene changes can modify the competence of a tissue. 

Comprehension of physiological genetics must, in turn, ultimately involve a detailed 
understanding of how genes influence metabolism, and since metabolism involves the 
study of physical and chemical reactions, phenogenetics must ultimately be described 
in biophysical and biochemical terms. The phenogenetic study of the gene causing 
sickle cell anemia proceeded from morphology to physiology to biochemistry. 



REFERENCES 

Dobzhansky, Th., and Holtz, A. M., "A Re-examination of Manifold Effects of Genes 
in Drosophila melanogaster," Genetics, 28:295-303, 1943. 

Goldschmidt, R. B., Theoretical Genetics, Berkeley and Los Angeles: University of 
California, 1955. 

Gluecksohn-Waelsch, S., "Physiological Genetics of the Mouse," Adv. in Genet., 4:2-49, 
1951. 

Griineberg, H., The Pathology of Development, Oxford: Blackwell, 1963. 

Hadorn, E., "Patterns of Development and Biochemical Pleiotropy," Cold Spring Harb. 
Sympos. Quant. Biol., 21:363-374, 1956. 

Hadorn, E., Developmental Genetics and Lethal Factors, New York: J. Wilev & Sons, 
1961. 

Kallman. F. J.. Heredity in Health and Mental Disorder, New York: Norton, 1953. 

Landauer, W.. "On the Chemical Production of Developmental Abnormalities," J. Cell 
Comp. Physiol.. 43 (Suppl.) : 26 1-305, 1954. 

Montagu. A.. Human Heredity, Cleveland: World, 1959. 

Newman. H. H.. Multiple Human Births. New York: Doubleday. Doran, 1940. 

Osborn. F.. Preface to Eugenics. Rev. Ed., New York: Harper, 1951. 

Osborn, R. H.. and De George, F. V., Genetic Basis of Morphological Variation. Cam- 
bridge, Mass.: Harvard University Press. 1959. 

Sang, J. H., "Penetrance, Expressivity and Thresholds," J. Heredity, 54:143-151, 1963. 

Waddington, C. H., New Patterns in Genetics and Development, New York: Columbia 
University Press, 1962. 

Wright. S.. "The Physiology of the Gene," Physiol. Rev., 41 :487-527. 1941. 



Phenotypic Effects <>i dene Action 



87 



Richard Benedict Goldschmidt 

(1878-1958). (From Genetics, 
vol. 45, p. I, 1960.) 



tf aH* * J) 




L 






.- 


?sji 






^^^J|& Wi 








k9 V? 






Hfr^sS^j 









QUESTIONS FOR DISCUSSION 

6.1. Can two genetically different individuals ever have identical viabilities? Explain. 

6.2. Why can you not conclude, from the evidence presented, that the genes for MN 
blood type in man, or for auria phenotypes in the snapdragon, are pleiotropic? 

6.3. How can genes be lethal to a genotype without producing a corpse? 

6.4. Two curly-winged, stubble-bristled Drosophila are mated. Among a large num- 
ber of adult progeny scored the ratio obtained is: 4 curly stubble: 2 curly only: 2 
stubble only:l neither curly nor stubble (therefore normal, wild-type). Explain 
these results genetically. 

6.5. In Drosophila, a mating of <5 A X 9 B or of S CX $ D produces F,, Vi of 
which turn brown and die in the egg stage. If, however, the matings are 
t! A X 9 D or <3 C X 2 B, none of the F, eggs turn brown and die. How can 
you explain these results genetically? 

6.6. In what respects are the terms penetrance and dominance similar and in what 
respects are they different? 

6.7. Is it the gene for dull red eye color that is pleiotropic in Drosophila, or is it the 
allele for white eye color? Explain. 

6.8. Most of the genes studied in Drosophila affect the exoskeleton of the fly. Do 
you suppose these genes also have effects on the internal organs? Why? 

6.9. Would you expect to find individuals who are homozygous for Polydactyly? 
Explain. What phenotype would you expect them to have? Why? 

6.10. Why are genes whose penetrance is 100% and expressivity is uniform particularly 
valuable in a study of gene properties? 

6.11. Two normal people marry and have a single child who is polydactylous on one 
hand only. How can you explain this? 



SS CHAPTER 6 

6.12. A certain type ol baldness is due to a gene thai is dominant in men and recessive 

in women. A nonhald man marries a bald woman and they have a bald son. 
Give the genotypes ol all individuals and discuss the penetrance of the genes 
involved. 

6.13. A man has one brown eye and one blue eye. Explain. 

6.14. How could you distinguish whether a given phenotype is due to a rare dominant 
gene with complete penetrance or a rare recessive gene of low penetrance? 

6.15. In determining whether or not twins are dizygotic, why must one study traits 

tor which one or both parents arc hctcrozygotes? 

6.16. Are mistakes ever made in classifying twins as dizygotic in origin? Why? 

6.17. Can the gene P for Polydactyly be considered as being partially dominant? As 
having pleiotropic effects? Explain. 

6.18. When nonidentical twins are discordant for ABO blood type, why must one or 
both parents have been heterozygous for /' or / /; ? 

6.19. Invent a particular situation that would result in greater discordance for identical 
than for nonidentical twins. 

6.20. What would be the probability of twins being dizygotic in origin if both have 
the genotype aa Bb CC Dd Ee Ff, each pair of alleles segregating independently, 
if the parents are genotypically Aa Bb CC DD Ee Ff and Aa BB CC dd ee FF? 

6.21. How would you test whether, in women, there is a genetic basis for the matura- 
tion of more than one egg at a time? 

6.22. In what way can you imagine that the paternal genotype could influence the 
frequency of twinning? 

6.23. Is tuberculosis "inherited"? Explain. 

6.24. What can twin studies by themselves tell you about genes? About genetic re- 
combination? 

6.25. Is it valid to apply the conclusions from twin studies to nontwin members of the 
population? Explain. 

6.26. Does this chapter present any new information about genetic properties? Explain. 

6.27. In a genetically black strain of the house mouse, W. L. Russell found a mouse 
with a splotchy phenotype — having white spotting on the belly and occasionally 
on the back. Splotchy X black gives both splotchy and black types of progeny. 
Splotchy X splotchy also produces the same types, but a number of embryos die 
in utero at 14 days of age and are characterized by a kinky tail and spina bifida. 
Discuss the genetic basis for and the dominance relationships involved in these 
results. 

6.28. It has been found that mouse ovaries transplanted from embryos to adult females 
can develop to maturity and produce offspring. Describe how you would proceed 
to determine the genotype of the abnormal embryos described in 6.27. 

6.29. Do you agree with J. H. Sang that penetrance (P) and expressivity (E) ". . . are 
descriptive terms which cloak our ignorance of the underlying reactions which 
determine particular values of P and E in any situation"? Explain. 

6.30. In the Japanese quail (Coturnix coturnix) matings between normal-appearing 
individuals of certain strains produce some micromelic embryos, having a short 
broad head with bulging eyes, which die between 11 and 16 days of incubation. 
How would you proceed to determine whether these abnormal embryos are 
homozygotes for a single recessive lethal gene? 



Phenotypic Effects of Gene Action 89 

6.31. What conclusions can you draw from the data of B. Harvald and M. Hauge 
(J. Amer. Med. Assoc, 186:749-753, 1963) obtained from an unbiased sample 
of Danish twins? 

One twin Both twins cancerous 

Twin pairs cancerous At same site At different sites 

Identical 1528 143 8 13 

Nonidentical 2609 292 9 39 

6.32. In what way does the study of genes help us understand normal embryonic 
development? 

6.33. If most somatic cells have the same genetic content, why do different cells not 
differentiate in the same way? 

6.34. In what ways can genes regulate embryonic development? 

6.35. Do the studies of Creeper, of Brachy, or of pituitary dwarfism in mice offer any 
support for the view that most, if not all, genes have a single, primary effect? 

6.36. What is the relationship between phenogenetics, developmental genetics, physio- 
logical genetics, and biochemical genetics? 

6.37. Discuss the comparative importance of genes that act earlier, as compared with 
those which act later, in development. 

6.38. Do you suppose that all genes act at all times in all cells of the body? Why? 

6.39. "This chapter tells more about development than it does about genes." Do you 
agree? Why? 

6.40. What can be learned about gene action if the gene studied ( 1 ) has only two 
alternatives, (2) has many alternatives? 



Chapter 7 

SEX CHROMOSOMES 
AND SEX-LINKED GENES 



S 



|i\( i sex has phenotypic alter- 
natives (maleness and female- 

ness ) . the genetie basis for sex 
can be investigated. This basis cannot be 
determined by studying garden pea plants 
because they are bisexual; that is. both sex- 
ual alternatives occur in one individual, and 
no phenotypic differences will be produced 
by genetic recombination. The typical 
Drosophila individual, however, being either 
male or female (Figure 2-6, p. 23), can be 
used to study the genetic basis for sex. 

When normal males and females mate to- 
gether, the male: female ratio of their prog- 
eny is approximately 1:1. This suggests the 
simplest hypothesis — that sex in Drosophila 
is determined by a single gene pair, one of 
the sexes being a homozygote and the other 
a heterozygote. For the moment, however. 



which sex corresponds to which genotype 
cannot be designated. 

In accordance with the view that chromo- 
somes contain the genes, one pair of chro- 
mosomes should be concerned with sex. Let 
us call the presumed homologous pair of 
chromosomes carried by the homozygote 
lor the sex genes the XX pair and those 
carried by the heterozygote, the XY pair. 
Segregation and random cross fertilization 
then would produce equal numbers of XX 
and XY progeny. The X and the Y chro- 
mosomes presumed to carry the genes for 
sex can be called sex chromosomes; each of 
the other chromosomes which an individual 
carries can be called an autosome (A). 
Since Drosophila melanogaster has a diploid 
chromosome number of four pairs, each 
individual can be represented as either 
XX + 3AA or XY + 3AA. 

Sex-Linked Genes 

Consider the results of certain crosses in- 
volving the recessives cubitus interruptus 
(ci) and ebony body color (e) and their 
dominant alleles ci + (normal wing venation) 
and e + (gray body color). One cross, 
ci+ci e + ehy ci ci e e — a dihybrid parent and 
a double recessive parent (Figure 7-1 ) — 
produces offspring in a 1:1:1:1 ratio thus 



+ + 
ci ci e e 



ci ci e e 



+ + + + 

'Aci e, V4ci e, V-ici e, 'Aci e 



ci e 



+ + 

Aci ci e e 

Aci ci e e 

'Aci ci e e 

'Aci ci e e 



FIGURE 7-1 . Results of hack- 
crossing a dihybrid. 



90 



,S>.\ Chromosomes and Sex-Linked denes 



\)\ 



demonstrating that the two pairs of genes 
are segregating independently. The same 
result and conclusions hold for the cross of 
ci ci e? by ci ci e * e. Consider, next, 
crosses in which the sex and wing venation 
traits are studied simultaneously in recip- 
rocal matings — ci+ci XX by ci ci XY and 
ci ci XY by ci ci XX. In both cases the 
result is a 1:1:1:1 ratio of cubitus male, 
cubitus female, normal male, normal female. 
Here, then, the sex genes segregate inde- 
pendently of the genes for cubitus. There- 
fore, according to our hypothesis, the ci 
alleles are located autosomally. 

Similarly, each of the reciprocal crosses — 
e fXXbyee XY and e • e X Y by e e XX 
— also gives a 1:1:1:1 ratio, indicating that 
the gene for ebony is located autosomally. 
Since the genes for ebony and cubitus segre- 
gate independently of each other, they must 
be located on different pairs of autosomes. 

Even though we cannot yet specify which 
sex is XX or XY, the last two types of 
crosses can be described as reciprocally made 
backcrosses of a monohybrid; that is, one 
time the hybrid parent was the male, the 
other time the hybrid parent was the female. 
In both cases the two traits appear in a 1:1 
ratio among the sons, and in a 1:1 ratio 
among the daughters. 

At this point, an earlier statement (p. 
32) — that all crosses give the same results 
when made reciprocally — can be understood 
to mean that the observed phenotypes 
and their proportions are the same for 
sons and daughters even though the crosses 
were made reciprocally. So, for example, 
in a cross of the dihybrids ci+ci e + e by 
ci+ci e + e there would be a 9:3:3:1 ratio 
among the sons and a 9 : 3 : 3 : 1 ratio among 
the daughters because the parents' sex genes 
were located in the sex chromosomes, where- 
as the other gene pairs were in nonhomolo- 
gous pairs of autosomes. Previously, all the 
crosses we dealt with involved autosomal 
recombination. Because autosomal genes 



A 



B 



P -'9* whi^ P "hi,e CJ*^ 



lei'? 9 



/hite 



dull 
red 



0*0* 

?? 



figure 7-2. Phenotypic results of reciprocal 
matings involving eye color. & 6 nudes. 
9 9 = females. 



always segregate independently of the sex 
genes, sex did not influence the results; that 
is, the phenotypic results of autosomal re- 
combination crosses are the same for sons 
and daughters even though reciprocal mat- 
ings are made. 

But consider the results of crosses involv- 
ing the dull-red (w+) and white (w) eye 
color alleles. Using pure lines, dull-red 9 
by white i (Figure 7-2 A) produces all 
dull-red sons and daughters in F,, as ex- 
pected, since w+ is dominant. However, 
the reciprocal cross (Figure 7-2B), white 
9 by dull-red 6 , produces only white sons 
and dull-red daughters. Although the first 
cross gives the same result for sons as for 
daughters, the second (reciprocal) cross 
gives different results: sons resemble their 
mothers; daughters resemble their fathers. 
Because such different results are never ob- 
tained from reciprocal matings involving 
autosomal genes, we can conclude that w+ 
and its alleles are not located autosomally. 

Let us assume that the gene for white eye 
is located in the sex chromosomes and, 
therefore, is sex-linked and investigate the 
consequences of this on the gene's transmis- 
sion relative to the sex phenotype. 1 

If we assume that females are XY and 
males XX, the first cross then is dull-red 
female X""Y"" by white male X"X"' (Fig- 
ure 7-3, A-l ), and the F, expected are 

1 See T. H. Morgan ( 1910). 



92 



CHAPTER 7 



A-l B-l 

9W W ^fl W W ("*) WW X 

xxx(j p, x y qp x X X (J 

W fl^^ WW fl ^ 

x OO F x x OCT 

+ 99 W99 



+ + 

W W ( 1 WW 

P, X Y x X X 



w 



w w 
X Y 



A-2 

+ + 
P, X X X : 



B-2 



Q w w 



+ + 



^ w wO w w 

Q P,XXqpxXY 



a* 



+ 
w w 

X Y 






w 


w 


X 


Y 


+ 


w 


w 


X 


X 



A-3 



99 

B-3 



+ + 



O WW ^_^ W W ("") WW ^ 

p,xx Y xXY U p, x x ; x x y Q 



w 



>99 



w w 
X Y 



0*0* 

' x->99 



figure 7-3. Three attempts (A-l and B-l, A-2 and B-2, A-3 and B-3) 
to represent matings A and B in Fig. 7-2 genotypically. Shaded genotypes 
must he incorrect. 



X"X" sons and X"Y"' daughters, all 
dull-red-eyed, as found. The reciprocal 
cross (Figure 7-3, B-l), therefore, is white 
9 X"Y" by dull-red $ X W+ X M '\ The 
Fi daughters (X"'*Y W ) are expected to be 
dull-red-eyed, as found. The Fi sons 
(X"*X" '), also expected to be dull-red-eyed, 



are, however, actually white-eyed. There- 
fore, we must reject this particular hypoth- 
esis for correlating sex chromosomes and eye 
color genes. 

So let us assume the reverse situation — 
that females are XX and males XY. The 
same crosses are represented now as dull-red 



Sex Chromosomes and Sex-Linked Genes 



93 



9 X w+ X w+ by white 6 X"Y' r producing 
X" + X"' (dull-red) daughters and X" + Y'" 
(dull-red) sons (Figure 7-3, A-2); recip- 
rocally, white 9 X"'X' r by dull-red $ 
X w+ Y w+ produces X w+ X w (dull-red) daugh- 
ters and X W Y W * (dull-red) sons (Figure 
7-3, B-2). Again an expected phenotype 
is contrary to fact — the phenotype of the Fi 
sons being white, not dull-red. 

Since we cannot explain these observa- 
tions by identifying maleness with XX or 
XY alone, we must increase the number of 
assumptions. Let us then test two hypoth- 
eses simultaneously: (1) Assuming that 
Drosophila males are XY and (2) the Y 
chromosome carries w but no other allele, 
then the genotypes and results of the first 
cross described in the preceding paragraph 
remain unchanged (Figure 7-3, A-3); the 
reciprocal cross (Figure 7-3, B-3) becomes 
white 9 X"X"" by dull-red <$ X M ' + Y\ 
producing X"X" (dull-red) daughters and 
X I( Y"" (white) sons. Since these hypoth- 
eses fit the observations, we may accept 
them. 

The recombination genetics of several 
traits in Drosophila other than sex and eye 
color also proves to be based upon a pair of 
genes in the sex chromosomes; that is. each 
case gives different results in Fi when lines 
pure for different alternatives are crossed 
reciprocally. Moreover, each case can be 
explained by assuming that females are XX, 
males XY, and that the Y carries the most 
recessive and least effective allele known for 
the gene pair under test, as is w in the case 
of eye color. In such cases, the absence 
from the Y of a partially or completely domi- 
nant allele must mean that such alleles can- 
not be produced by mutation of the most 
recessive allele simply because this recessive 
allele does not exist on the Y. Accordingly, 
the Y ordinarily lacks an allele of a gene 
located on the X; therefore, in Figure 7-3, 
A-3 and B-3 a Y should be substituted 
for each Y" . 



Whenever the Y carries no allele of a gene 
on the X, sons will express phenotypically 
whatever allele is contained in the single X 
each son receives from his mother. With 
regard to these genes, therefore, a Drosoph- 
ila female is being test crossed whenever 
and to whomever she mates, since her X 
chromosome genotype can be determined di- 
rectly from the phenotypes of her sons. An 
otherwise diploid individual carrying one or 
more unpaired genes is said to be hemi- 
zygous in this respect. For example, a gene 
in the X chromosome with no allele in the 
Y is hemizygous in the Drosophila male; 
half of the zygotes he produces will receive 
this allele in the X he contributes, whereas 
the half receiving the Y will not get one. 
The X of a Drosophila male is obtained 
from his mother and transmitted to each of 
his daughters; the Y is transmitted from 
father to son. 

In poultry a mating of a female with non- 
barred feathers to a male with barred feathers 
produces offspring which are all barred — 
barred (B) being dominant to nonbarred 
(b) ( Figure 7-4 A ) . In the reciprocal cross 
(Figure 7-4), barred 9 by nonbarred $, 
all sons are barred and all daughters non- 
barred. Here also the results of reciprocal 
matings differ, so that we are dealing again 
with sex-linkage. In the reciprocal cross, 
note that the exceptional F x are nonbarred, 
showing the recessive trait as in the case of 
Drosophila. But, in poultry the sex is op- 
posite, since the exceptional Fi are females. 
(The exceptional F! Drosophila were white- 
eyed males.) To explain these results we 
must assume that in poultry, as in Drosoph- 
ila, sex is determined by XX vs. XY, and 
that the X chromosome does and the Y 
chromosome does not contain a gene for 
barred or nonbarred feathers. But, contrary 
to Drosophila, poultry males are XX and 
females, XY. 

The genotypes of the bird crosses are, on 
these hypotheses, X 6 Y (nonbarred 9 ) by 



94 



< HAPTER 7 



\ V (barred >, producing X B Y (barred 
9 ) and X'.V (barred ) in F,; the re- 
ciprocal mating of .VY (barred V ) by 
\ \ (nonbarred ; ) produces VY (non- 
barred 5 ) and V V (barred ) in F, 
( Figure 7-4. A-l and B-l ). 

Support for die existence ol sex chromo- 
somes may be sought from etiological ob- 
servations. If the gene content of the X and 
> is different as in Drosophila and poultry. 
the cytological appearance of the two kinds 
oi sex chromosomes might also be different, 
i Note, however, that the preceding explana- 
tion of sex-linkage was made independently 
of any cytological expectation.) 

In Drosophila it is found cytologically 
(Figure 7-5) that three of the four pairs of 
chromosomes seen at mitotic metaphase cor- 



respond in the male and the female, homo- 
logs being \er\ similar morphologically. In 
the female the homologs of the fourth pair 
are also morphologically similar. In the 
male, however, only one member of this 
pair looks like its homologs in the female; 
its partner's morphology is distinctly differ- 
ent. Thus, the distinctive cytological ap- 
pearance of this last chromosome is con- 
sistent with the genetic expectation for a Y 
chromosome being present once in the male 
and not at all in the female. The other 
homolog in the male is then called the X and 
is present twice in the female. The reverse 
cytological picture is observed in poultry; 
here the homologs are similar for each pair 
of chromosomes in the male, whereas the 
female has one heteromorphic pair; that is, 



A 



Nonbarred ^ 

Barred f~Y?~) 
Barred QO 



x Barred 



B 

/-V^ P, Barred V/ x Nonbarred r^ 

a"a" 
99 



Barred 



Nonbarred 



b 
X Y 



A- 



V x X X (J 



B 
X Y 



2 



B-l 



b b * 

X x O 



B b 
X X 



x%99 



B b 
X X 



X Y 



do' 
99 



FIGURE 7-4. Phenotypic (A and li) and genotypic (A-J and B-l) results of reciprocal 
nuttings involving barred and nonbarred feathers in chickens. 



Sex Chromosomes and Sex-Linked denes 



95 



FEMALE 



MALE 



A tf 

xx * 

figure 7-5. Silhouettes of chromosomes of 

Drosophila melanogaster as seen at mitotic 
metaphase. 



one pair whose members are morphologically 
different, one being similar to, one different 
from, the corresponding pair in the male. 

As in birds, the male of moths, butter- 
flies, and some amphibians and reptiles is 
XX and the females, XY. In human beings, 
genetic and cytological evidence shows XY 
to be male and XX to be female, just as in 
Drosophila. So in different species different 
sexes make two kinds of gametes; that is, 
different sexes are heterogametic with re- 
spect to sex chromosomes. 

In man, a certain kind of red-green color- 
blindness is sex-linked due to a recessive 
allele, c, present on the X and absent on 
the Y. Accordingly, color-blind women 
(X'X C ) who marry normal men (X r Y) have 
normal daughters (X r X r ) and color-blind 



figure 7-6. Pedigree showing a woman homo- 
zygous for the gene for hemophilia. 



m^5~i 



r 



6 



b 



sons (X'Y). The classical bleeder's disease 
in human beings, hemophilia type A, is also 
due to an X-linked recessive gene, h, absent 
from the Y. This rare disease usually occurs 
in males. Recently, however, a few hemo- 
philic women have been discovered in Eng- 
land. These homozygotes are extremely rare 
because, barring mutation, they must have a 
hemophilic father (X'Y) and a heterozy- 
gous mother (X"X*) — (Figure 7-6). 



A. PHENOTYPES 



P, White O x DuM w-* 



B. GENOTYPES 



WW w 

X X x X Y 



F, TYPICAL 



EXCEPTIONAL 




X Y 



w w 
X X 



H 

[w wl 
X xj 



figure 7-7. Nonmutant exceptions in crosses 
involving eye color in Drosophila. 



Nondisjunction 

Certain additional experiments have been 
performed with the sex-linked gene for white 
eye in Drosophila.- When white females 
(X"X") are crossed with dull-red males 
(X" Y) and large numbers of progeny are 
scored, not all F ( are white sons (X"Y) 
and dull-red daughters (X w+ X w ) as ex- 
pected according to sex-linkage. One or 
two flies per thousand F t are exceptional 
dull-red-eyed sons or white-eyed daughters 
(Figure 7-7). These exceptional flies can- 
not be explained away as the result of care- 
less scoring of phenotypes or contamination. 
Moreover, they cannot be explained as being 
due to mutation, since the mutation frc- 

- Based upon work of C. B. Bridges. 



96 



CHAPTER 7 



quencj from w to w or the reverse is 
several orders of magnitude lower than the 
observed frequency of exceptional Hies. 

Since the exceptional Fi females are white- 
eyed, each must cam \ V | Figure 7-7B). 
The onlj source of X's containing w is the 
mother. Accordingly, the father must fail 
to contribute his X"' chromosome to an 
exceptional daughter. Each exceptional 
dull-red-eyed son must carry X"", which 
could be contributed only by the father. 

In order to understand how this excep- 
tional situation may come about, let us 
examine the normal consequence of meiosis 
in the Drosophila female as regards the sex 
chromosomes. Normally, the two X's syn- 
apse and form a tetrad, and due to segrega- 
tion four nuclei are produced at the end of 
meiosis, each containing one X (Figure 
7-8 A ) . One of the four nuclei becomes the 
gametic (egg) nucleus; the other three are 
discarded (in polar bodies). 

Suppose, however, segregation of the four 
strands in the X chromosome tetrad occa- 
sionally occurs improperly in either of two 
ways: 

1. At anaphase I, instead of one dyad 
going to each pole, both dyads go to the 
same pole (Figure 7-8B). The daughter 
nucleus containing no X dyad undergoes the 
second meiotic division to produce two 
nuclei, neither one having an X. The other 
daughter nucleus, containing two dyads, pro- 
ceeds through the second division, during 
which the two members of each dyad sepa- 
rate and go to opposite poles at anaphase II. 
The result is two daughter nuclei each con- 
taining two X's, one from each dyad. 
Therefore, at the end of meiosis, the failure 
of dyads to disjoin at anaphase I will result 
ultimately in four nuclei, two with no X and 
two with two X's. As a consequence the 
nucleus which becomes the gametic nucleus 
has a 50% chance of carrying no X and a 
50% chance of carrying two X's. 

2. Alternatively (see Figure 7-8C), 



METAPHASE TELOPHASE 




figure 7-8. Consequences of normal segrega- 
tion of X chromosomes (A) and of its failure 
to occur (B and C) . 



anaphase I is normal, and at telophase I 
two daughter nuclei are formed each con- 
taining one X dyad. The second meiotic 
division occurs normally in one of the daugh- 
ter nuclei, producing two telophase II nuclei, 
each of which contains one X. In the other 
daughter nucleus, however, the members of 
the X dyad fail to separate at anaphase II 
and go instead into the same telophase II 
nucleus. This failure of monads to disjoin 
at anaphase II produces two nuclei, one con- 
taining no X and the other containing two 
X's. Consequently, the gametic nucleus has 
a 25% chance of carrying no X, a 25% 
chance of carrying two X's, and a 50% 
chance of carrying one X. 

The occasional failure of normal separa- 
tion of chromatids at either the first or the 
second meiotic division would result in the 
occasional production of eggs containing 
either no X or two X's. Such a failure of 
the members of a pair of chromosomes to 
segregate is also referred to as nondisjunc- 
tion of chromosomes. According to the hy- 
pothesis that the X chromosome carries an 
allele for w, chromosomal nondisjunction 
can provide the mechanism by which a pair 
of genes fails to segregate, with the result 
that after meiosis, eggs are sometimes pro- 
duced containing two members or no mem- 



Sex Chromosomes and Sex-Linked Genes 



97 



bers of the gene pair. Any egg produced 
following nondisjunction will usually be fer- 
tilized by a sperm carrying either an X or 
a Y in addition to a haploid set of auto- 
somes. (Nondisjunction can also occur dur- 
ing meiosis in the male. We can ignore 
this complication here, because it is an in- 
frequent event and the probability that an 
egg produced after nondisjunction would be 
fertilized by a sperm produced after nondis- 
junction is negligible.) 

If the hypothesis of chromosomal nondis- 
junction is valid, it should be consistent with 
the genetic results. After nondisjunction the 
exceptional eggs produced by a white 
(X"X ,r ) female would be either X W X W or 
(zero designating the absence of the 
homolog normally expected to be present). 
The normal sperm produced by a dull-red 
(X' t+ Y) male would carry either X w+ or 
Y. The expected genotypes of Fi following 
random fertilizations between these gametes 
are given in Figure 7-9. 

Let us momentarily ignore the sex of these 
exceptional offspring and classify them only 
for eye color. Type 1 would be dull-red- 
eyed, type 2 white-eyed, type 3 dull-red- 
eyed, and type 4's eye color undetermined. 
The genetic observations would be explained 
if types 1 and 4 were lethal; type 2, female; 
type 3, male. (On the hypothesis that XX 
is female and XY is male, it is reasonable 
to assume that types 1 and 4 would be 
neither, and therefore might be lethal. ) 
Even more specific requirements must be 
fulfilled before accepting these hypotheses, 
namely, that each exceptional white female 
must prove to be XXY cytologically; that is, 
such females must have, in addition to the 
normal diploid chromosomes of a female, 
an extra chromosome which is Y. More- 
over, each exceptional male must have, in 
addition to the normal autosomes, one X 
but no Y. When the somatic cells of excep- 
tional females and males are examined cyto- 
logically, these chromosomal prescriptions 



are found to be filled completely. It is also 
possible to show that YO zygotes are lethal, 
and that X"*X"X" individuals— these usu- 
ally die before adulthood — are dull-red-eyed. 
Although XY individuals are fertile males, 
XO flies are invariably sterile males. This, 
therefore, implies that the Y chromosome is 
necessary for male fertility, the trait being 
attributable to a gene on the Y which has 
no allele on the X. Moreover, our sex chro- 
mosome formula for maleness must be mod- 
ified to include XY and XO individuals and 
similarly the femaleness formula also modi- 
fied to include XX and XXY individuals. 

Chromosomes as Genetic Material 

We should now re-evaluate the hypothesis 
that the chromosomes serve as the material 
basis for genes. In preceding chapters, the 
following parallels were found between the 
properties and behavior of genes and of 
chromosomes: both come only from pre- 
existing counterparts; both are self-replicat- 
ing; both occur as pairs in all cells of the 
diploid stage of sexually reproducing organ- 
isms except gametes; both are replicated in 
each mitotic division; both maintain their 
individuality from one mitotic division to the 



EGGS 


SPERM 


OFFSPRING 


w w 
X X 


+ 
w 

X 


(1) 


+ 
www 

XXX 


w w 
X X 


Y 


(2) 


w w 
XXY 





+ 
w 

X 


(3) 


+ 
w 

X 



Y (4) Y 

figure 7-9. Genotypic expectation after fer- 
tilization of nondisjunctionally produced eggs 
by normal sperm. 



98 



CHAPTER 7 



next; both arc capable of imitation and sub- 
sequent replication of the new form; both 
segregate during gametogenesis so that they 

occur unpaired in the gamete; both show 
independent segregation for different pairs; 
both are combined randomly at fertilization. 
It was also hypothesized that the chromo- 
some is larger than a rccombinational gene 
(the smallest rccombinational unit of the 
genetic material ). since a gene was described 
as the largest distance along the length of a 
chromosome within which an exchange lead- 
ing to a chiasma cannot form. 

These parallels still might be considered 
merely coincidental. The present chapter 
provides the following additional tests of the 
idea that chromosomes function as the ma- 
terial basis for genes: 

Se.x-linka^e. detected by the nonrandom 
association between sex and the genes for 
certain traits, was found to be an exception 
to the mode of transmission of the auto- 
somal genes studied previously. This phe- 
nomenon was explained only by the assump- 
tion that certain genes have an allele in the 
homologous chromosome of a pair in one 
sex but no allele in the homologous chromo- 
some in the other sex. Hemizygosity was a 
necessary assumption in the case of the 
Drosophila male and the poultry female. 
This genetic aberration was exactly paral- 
leled by the occurrence of a pair of hetero- 
morphic homologs in the Drosophila male 
and poultry female, one homologous mem- 
ber being present as a pair in the female 
Drosophila and in the male chicken. 

Finally, in Drosophila, an exception to the 
exception of sex-linkage was found and ex- 
plained genetically as resulting from the 
failure of the members of a single pair of 
sex-linked genes to segregate. This failure 
produces gametes containing two or no 
alleles of a given sex-linked gene. This 
genetic nondisjunction was shown to result 
from chromosomal nondisjunction; that is, 




Calvin Blackman Bridgi s (1889-1938). 
(From Genetics, vol. 25. p. 1. 1940.) 

the members of a pair of X chromosomes 
failed to segregate during meiosis. From 
chromosomal nondisjunction, it was pre- 
dicted that the different genetically excep- 
tional individuals would have different spe- 
cific and unique sex chromosomal composi- 
tions, and further tests proved this was the 
case. 

In the light of these results, the view that 
the chromosomes are the material basis for 
all the genes so jar studied can no longer be 
considered merely a hypothesis based upon 
limited — therefore possibly circumstantial — 
evidence, but must now be accepted as a 
theory supported both by all the typical and 
all the atypical recombinational properties 
of genes and of chromosomes. 

Ordinarily, no further comment will be 
made in this book about new tests that sub- 
stantiate the theory. Henceforth, assume 
that all tests do so unless note is made to 
the contrary. 



Sex Chromosomes and Sex-Linked Genes 



SUMMARY AND CONCLUSIONS 



99 



Up to now we have studied genes located in autosomes. We have found that the 
recombination of autosomal genes is such that reciprocal crosses between different pure 
lines produce F, which are genotypically and phenotypically uniform; that is. there 
is no dependency between the traits which appear and the sex of the offspring because 
autosomal genes segregate independently of the genetic material in sex chromosomes. 

In Drosophila, sex is not the only trait directly associated with genetic material 
located in the sex chromosomes. Several other traits in Drosophila yield results which 
differ in reciprocal matings made between lines that are pure for different alternatives: 
the difference appearing in the phenotypes shown by one of the sexes. Genes behaving 
in such a sex-linked way are not located autosomally. The Y sex chromosome carries 
no allele of these genes, and the X does. 

In human beings and Drosophila, the XY sex chromosomal constitution is male and 
the XX female; in birds and moths, it is the female which is heteromorphic, and, 
therefore, heterogametic with reference to sex chromosomes. 

Occasionally, as a consequence of nondisjunction of sex chromosomes at meiosis, 
chromosome segregation fails, and gametes are formed containing two or, comple- 
mentarily, no sex chromosomes. When this nondisjunction occurs in a Drosophila 
female homozygous for an X-linked recessive gene and such a female is mated to a 
male carrying the dominant allele, some offspring appear that are simultaneously ex- 
ceptions to sex-linkage and to sex chromosome content; the exceptional feature of the 
one accurately predicts the exceptional feature of the other, and vice versa. 

Sex-linkage and nondisjunction offer additional tests of the hypothesis that the ma- 
terial basis of all the genes studied thus far is in the chromosomes. This view is sup- 
ported by so many and diverse lines of evidence, and contradicted by none, that it 
must be accepted as theory. 

REFERENCES 

Bridges, C. B., '"Non-Disjunction as Proof of the Chromosome Theory of Heredity " 
Genetics, 1:1-52, 107-163, 1916. 

Morgan. T. H., "Sex Limited Inheritance in Drosophila," Science, 32:120-122, 1910. 
Reprinted in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood' Cliffs 
N.J.: Prentice-Hall, 1959, pp. 63-66. 



QUESTIONS FOR DISCUSSION 

7.1. Under what circumstances would sons fail to receive a Y chromosome from 
their father? 

7.2. In the cross X""Y by X"X" what would you expect to be the genotypes of the 
zygotes produced, after sex chromosome nondisjunction during meiosis in both 
male and female Drosophila? What is the phenotypic outcome in each case? 

7.3. If a trait is found to be due to a gene unlinked to any autosome, does this prove 
that the gene is linked to a sex chromosome? Explain. 

7.4. A husband and wife both have normal vision, although both their fathers are 
red-green color-blind. What is the chance that their first child will be: 

(a) a normal son? (c) a red-green color-blind son? 

(b) a normal daughter? (d) a red-green color-blind daughter? 



100 CHAPTER 7 

7.5. One child is hemophilic, his twin hrother is not. 

(a) What is the probable sex of the hemophilic twin? 

(h) Are the twins monozygotic? Explain. 

(c) Give the genotypes of both twins and of their mother. 

7.6. A hemophilic father has a hemophilic son. Give the most probable genotypes 
oi the parents and child. 

7.7. A Drosophila male with cubitus interruptus wing venation, ebony body color, 
and white eye color is mated to a pure wild-type female (normal wing venation, 
gray body color, and dull-red eyes); then the F, females are crossed to males 
like their father. Give the kinds and relative frequencies of genotypes and of 
phenotypes expected among the offspring of the last cross. 

7.S. Are you convinced that all genes have their material basis in the chromosomes? 
Explain. 

7.9. What reason can you give for believing that in Drosophila the Y chromosome is 
lacking a gene present in the X chromosome? That the X is lacking a gene 
present in the Y? 

7.10. List evidence in support of the theory that chromosomes contain the material 
basis for genes. 

7.1 1. Has any evidence been presented that a chromosome carries more than one gene? 
Explain. 

7.12. What proportion of all genes causing hemophilia type A is found in human males? 
Justify your answer. 

7.13. Does a gene have to be hemizygous in one sex to be sex-linked? Explain. 

7.14. Two phenotypically wild-type Drosophila were mated in a vial. By accident all 
but one F x was lost. The survivor was a male with white eyes, ebony body 
color, and cubitus interruptus venation. Give the most probable genotypes of 
the parents. 

7.15. Using pure stocks of Drosophila, yellow-bodied male by gray-bodied (wild-type) 
female produced 1241 gray-bodied daughters, 1150 gray-bodied sons, and 2 
yellow-bodied sons. The reciprocal mating produced 1315 gray daughters, 924 
yellow sons, and 1 yellow daughter. Give the genetic and chromosomal makeup 
of each type of individual mentioned. Discuss the relative viability and fertility 
of the different chromosomal types. 

7.16. Females of Drosophila having a notch in their wing margins mated to wild-type 
males gave the following F, results: 550 wild-type 9 9, 472 notch 9 9, 515 
wild-type 6 6 . Explain these results genetically. 

7.17. A line of Drosophila pure for the sex-linked gene, coral (W") was maintained in 
the laboratory for many generations. To demonstrate sex-linkage to a class, a 
coral male was mated to a wild-type female, and all the F^ were as expected. 
The reciprocal cross, between a coral female and a wild-type male, gave 62 coral 
females and 59 wild-type males. Present a hypothesis to explain this unusual 
result. How would you test your hypothesis? 

7.18. The wild-type eye shape in Drosophila is ovoid. A certain mutant, X, narrows 
the eye. Using pure lines, and ignoring rare exceptions, mutant 9 X wild-type $ 
produces mutant sons and daughters in ¥ x ; wild-type 9 X mutant S produces 
wild-type sons and mutant daughters in Fj. 

Another mutant. Y, also narrows the eye. Using pure lines of Y and wild-type, 
mutant 6 or 9 X wild-type 9 or 2 produces 2 mutant 6 $ and 9 9:1 wild-type 
£ $ and 9 9. Discuss the genetics of mutants X and Y. 



Sex Chromosomes and Sex-Linked Genes 



101 



7.19. Reciprocal matings using pure lines of Drosophila produce all wild-type F, from 
wild-type by vestigial wings. What are the genotypic and phenotypic expecta- 
tions if the F a $ of this mating is crossed with a white-eyed vestigial-winged 9 
w being sex-linked? s ' 

7.20. Give two ways in which knowledge of sex-linked genes could be put to practical 
use. r 

7.21. A normal man of blood type AB marries a normal woman of O blood type 
whose father was hemophilic. What phenotypes should this couple expect in 
their children and in what relative frequencies? 

7.22. The diagram below is a partial pedigree of the descendants of Queen Victoria 
of England (II) which contains information regarding hemophilia only for gen- 
eration IV. In this generation, the entire symbol is filled in if the person has 
hemophilia. A heterozygote for hemophilia would have been represented by a 
half filled-in symbol. Fill in the symbols of previous generations using this 
system. & 



111 = Princess Alice 

112 = Leopold, Duke of Albany 

1111 = Irene 

1112 = Alexandra 

1113 = Alice 

III5 = Victoria Eugenie 



I VI = Prince Waldemar of Prussia 
IV3 = Prince Henry of Prussia 
IV8 = Tsarevitch Alexis of Russia 
IV10 = Viscount Trematon 
IV12 = Alfonso 
I VI 7 = Gonzalo 



II 



IV 



HI ( )1 



-n. 




oooo 



12 3 4 5 6 7 8 




OBO 

9 10 11 



12 13 14 15 16 17 

(After J. B. S. Haldane.) 



( 'hapter *8 

SEX DETERMINATION 



Drosophila 

In Chapter 7. it was mentioned that the 
ordinary Drosophila melanogaster female is 
3AA + XX and the male. 3AA + X + Y. 
One cannot decide the chromosomal basis 
for sex determination from these facts alone, 
however, since two variables are involved, 
the X's and the Y. Is the male a male be- 
cause he has a Y, because he has only one 
X, or because he has both one X and one 
Y? 

By knowing the sex of flies that carry — 
besides two sets of autosomes — either XXY 
(female). XX YY (female), or XO (male), 
we can see that the Y is not sex determining 
in this organism. (As has been indicated 
in Chapter 7, the Y is necessary for fertility. 
XO males having nonmotile sperm. ) 

Knowing that sex in Drosophila is cor- 
related with the chromosomal alternatives of 
XX versus X, one can ask: What is the 
detailed genetic basis for sex in terms of 
genes located in the X chromosome? The 



data so far presented can be interpreted to 
mean that only a single pair of genes (in the 
case of XX) or a single gene (in the case 
of X ) is the total genetic basis for sex de- 
termination. There are several implications 
in this interpretation. The X-linked sex 
gene need not have an alternative allele if 
the presence of one such gene produces one 
sex, and the presence of two, the other sex — 
dominance not being involved. It can also 
be claimed that the Y carries no allele for 
this sex gene. Two additional assumptions 
must be made, however, in Drosophila and 
in other species having heteromorphic sex 
chromosomes, to correlate the genetics with 
the cytology of sex. 

1. That the sex gene must be located in 
a region of the X which distinguishes 
X from Y cytologically. 

2. No chiasma may occur between X and 
Y within this cytologically different 
segment. 

These postulates are necessary to preserve 
the exact correspondence between the mor- 
phology of the X and its sex gene content. 
Consequently, even though a chiasma occurs 
between the X and Y in a segment which 
they share (for example, both carry an 
allele of bobbed), the resultant strands that 
appear cytologically as X will carry the sex 
gene, whereas those that appear as Y will 
not. These requirements are reasonable 



A. PHENOTYPiC RESULTS 



B. GENOTYPIC EXPLANATION 



Female x Male 


P, 


XX 


tra tra x 


XY tra tra 






G, 


'AX trat 'AX tra 


VjX tra, 'AY tra 






F, 
V 


, 25% 


XY tra tra 


d* 






/ 


25% 


XY tra tra 


a* 






\ 




\ 


> 25% 


XX tra tra 


(_} (transformed ^p) 




O r n Pii.iiibIiii 




• 25% 


XX tra tra 


? 


figure 8-1. Abnormal sex 
ratio in Drosophila. 


/d o Females 




102 













Sex Determination 



103 




since synapsis does not occur between non- 
corresponding regions of homologous chro- 
mosomes, and, in the absence of pairing, 
exchanges leading to chiasmata cannot 
occur. 

To further establish the cytogenetic basis 
for sex, we shall consider the results of 
crosses between certain laboratory strains of 
D. melanogaster. 1 One strain produces 
about 75% males and 25% females (Figure 
8-1 A), instead of the normal sex ratio of 
approximately 50% males and 50% females. 
Since just as many eggs become adult in 
this unusual strain as in a normal one, the 
abnormal result cannot be due to a gene 
that affects the viability of one sex. 

In this exceptional case it can be hy- 
pothesized that an autosomal gene is affect- 
ing the determination of sex. This gene is 
called transformer and is postulated to have 
two alleles, tra + , and tra. Homozygotes for 
tra are presumed always to form males re- 
gardless of the X genes present ( tra tra is 
epistatic and the X genes hypostatic), - 
whereas heterozygotes or homozygotes for 
tra + have their sex determined by the pres- 
ence of the sex gene on the X (in this case 
the X sex gene is epistatic). Accordingly, 
XX individuals that are also tra tra will ap- 

1 Based upon work of A. H. Sturtevant. 

-These terms are defined in Chapter 4. pp. 51-52. 




figure 8-2. Some abnormal sex types in 
Drosophila: A = superfemale; B = supermale; 
C = intersex. {Drawn by E. M. Wallace.) 
Compare with normal male and female in Fig. 
2-6. p. 23. 



pear as males {transformed females), ex- 
plaining the excess number of males in the 
progeny. Thus, a cross of XY tra tra (male ) 
by XXtra+ tra (female) (Figure 8-1B) 
produces one-fourth each XY tra tra (males), 
XY tra+ tra (males), XX tra tra (males, 
transformed females), XXtra+ tra (fe- 
males) — accounting for the numerical re- 
sults. All these assumptions have been 
tested in additional crosses and are con- 
firmed, proving that autosomal genes are 
also concerned with sex determination. 
Note, however, that the tra allele is very 



104 



CHAPTER 8 



rare: almosl all Drosophila found in nature 
arc homozygous tra 

So far we have described only two sex 
types in Drosophila. Occasionally, how- 
ever, individuals occur which have, overall, 
an intermediate sexual appearance: that is. 
they arc both male and female in certain 
respects. Such sexual intermediates, called 
intersexes (see Figure 8-2). are sterile. 
Intersexes arc relatively frequent among the 
progeny oi' triploid (3N) females (whose 
chromosomes at mitotic mctaphase are dia- 
gramed in Figure 8-3; X chromosomes arc 
represented by filled-in blocks, autosomes 
h\ blanks, and the Y by a broken line). 

Some o\ the gametes of triploid females 
are haploid and some diploid; still others 
contain one. two, or three nonhomologs with 
or without a haploid set. Whereas haploid 



eggs produce normal males and females 
when fertilized by sperm from a normal 
male, diploid eggs produce triploid females 
when fertilized bj X-bcaring sperm. Diploid 
eggs produce XX Y individuals with three 
sets o\ autosomes, however, when fertilized 
by Y-bearing sperm. Some intersexes have 
this chromosomal constitution; other inter- 
sexes carry three autosomal sets and XX — 
one X derived from an egg containing two 
autosomal sets and the other from an X- 
bearing sperm. 

Close observation reveals two additional 
sex types among the progeny of triploid 
Drosophila (Figures 8-2, 8-3). These do 
not appear as intersexes but as sterile "super- 
sexes" — one type, called a superfemale, 
shows characteristic female traits even more 
strongly than does the normal female, the 




SUPERFEMALE 



i^^fl !&'??■>■ 






SUPERMALE 




FEMALE 



INTERSEX 



MALE 



FIGURE 8-3. Chromosomal complements of the sexual types found among 
the progeny of triploid females of D. melanogaster. 



Sex Determination 



105 



PHENOTYPES 



figure 8-4. Sex index and 
sexual type in D. mclano- 
gaster. 



other type, supermale, shows characteristic 
male traits even more strongly than does 
the normal male. Chromosomally, the 
superfemale contains two sets of autosomes 
and three X's; the X's are derived from an 
egg which carries one set of autosomes plus 
XX and is fertilized by an X-carrying sperm. 
The superfemale usually dies before adult- 
hood (see p. 97). The supermale con- 
tains three sets of autosomes plus XY; the 
chromosomes are derived from an egg which 
carries two sets of autosomes plus X and is 
fertilized by a Y-bearing sperm. 

What conclusions can we draw about sex 
determination from a knowledge of the chro- 
mosomal composition of different sex types 
in Drosophila? 3 Since we know that genes 
in the X and in the autosomes are sex-deter- 
mining, let us refer to Figure 8-4, which 
tabulates the number of X's and sets of auto- 
somes present for each sex type and also the 
ratio of X's to sets of autosomes — a numer- 
ical sex index. This index ranges from 0.33 
for supermales to 1.5 for superfemales. 
Note that an index of 0.50 makes for male 
and that adding a set of autosomes can be 



3 The following is based upon work of C. B. 
Bridges. 



Superfemale 


3 




" tetraploid 


4 


Normal . 


triploid 


3 


Female 


diploid 


2 




^ haploid 


1 


Intersex 


2 


Normal male 


1 


Superma 


e 


1 



NO. X NO. SETS OF 

CHROMOSOMES AUTOSOMES 

(A sets) 



SEX INDEX 

No. X's 
No. A sets 

1.5 

1.0 

1.0 

1.0 

1.0 

0.67 

0.50 

0.33 



interpreted as creating more maleness, pro- 
ducing the supermale. When the sex index 
is 1 .0, essentially normal females are pro- 
duced, indicating that the female tendency 
of one X overpowers the male tendency of 
one set of autosomes. But if the index is 
between 0.50 and 1.00, intersexes are pro- 
duced, indicating, by the same line of rea- 
soning, that the effect of two X's is partially 
overpowered by the extra autosomal set 
present. Finally, when the sex index is 1.5, 
the female tendency of the X's becomes so 
strong that superfemales result. 

These results strongly suggest that sex de- 
termination is due to the balance of genes 
located in the X on the one hand, in the 
autosomes on the other. According to this 
view, only the balance of the genes involved 
is important, so that a sex index of 1 .0 
should (and does) produce a typical female, 
whether the individual is diploid (2X + 2 
sets of A), triploid (3X -f 3 sets of A), or 
tetraploid (4X -f 4 sets of A). Individuals 
that contain haploid (IX -f- 1A set) sections 
have been found and, as expected from their 
sex index of 1.0, these parts were female. 
Since all known facts support the exact cor- 
respondence between chromosomal constitu- 
tion and sexual types, we can accept chromo- 



106 



CHAPTER 8 



some balance as the typical basis of sex de- 
termination in Drosophila. 

What is the relationship between \-auto- 
some balance and tra, the sex-transforming 

gene'.' Sex is determined by X-autOSome 
balance when the individuals carry tra + , 
which they normally do. When tra is homo- 
zygous, however, the balance view does not 
apply and 2X + 2A sets produces a male. 

Gynandromorphs 

On relatively rare occasions, abnormal Dro- 
sophila appear with some of their parts 
typically male and the remainder, typically 
female. Such individuals are said to be 
mosaic for sex traits; sex mosaics are also 
called gynandromorphs or gynanders (Fig- 
ure 8-5). The male and female parts are 
clearly demarcated in such flies, sometimes 
the front and hind halves, at other times 
the right and left sides are of different sex. 
The sharp borderline between male and 
female parts in an insect gynander is due 
to the relatively small role that hormones 
play in insect differentiation, so that each 



figure 8-5. D. melanogaster gynandromorph 

whose left side is female and right side is male. 
(Drawn by E. M. Wallace.) 




body part is formed according to the geno- 
type it contains. In view of the preceding 
discussion, one would predict that the dip- 
loid cells in the female part of a gynander 
contain XX and those in the male part X, 
the chromosome number being otherwise 
normal. If this prediction is correct, then 
approximately half-and-half gynanders could 
originate as follows: the individual starts as 
a zygote containing 3AA -+- XX — that is, as 
a female. The first mitotic division of the 
zygotic nucleus is abnormal — one daughter 
nucleus contains 3AA + XX and is nor- 
mal, the other daughter nucleus contains 
3AA + X and is defective, because one of 
the X's failed to be included in this nucleus, 
degenerated, and was lost. However, sub- 
sequent nuclear divisions are normal — 
cells produced following mitosis of the XX 
nucleus and its descendants giving rise to 
female tissue, and cells derived from the X 
nucleus giving rise to male parts. In this 
case the gynander has about half its body 
male and half female. If, however, the X is 
lost at some later mitosis, a correspondingly 
smaller portion of the body will be male, 
explaining gynanders one quarter or less 
male. 

We can test whether this explanation is 
sometimes correct by making use of an X- 
linked gene which produces a phenotypic 
effect over a large portion of the body sur- 
face; that is, a gene that affects the size and 
shape of the bristles and hairs. Such a gene 
is forked, two of its mutant alleles being 
/ 346 and /. In homozygotes (females) and 
hemizygotes (males), / :: "' produces bristles 
and hairs of normal length and shape; / 
causes them to be shortened, split, and 
gnarled. The f' ih f heterozygotes have 
bristles and hairs slightly abnormal in these 
respects, showing a "weak forked" pheno- 
type. If a cross is made to produce female 
offspring that are f" ib /f heterozygotes, the 
following predictions can be made regarding 
the phenotype of the gynanders occasionally 



Sex Determination 



107 



present among the siblings: All gynanders, 
originating as postulated, will be weakly 
forked in their female parts; their male parts 
will have either normal or strongly forked 
bristles and hairs, depending upon whether 
the lost X carried / or f :1 \ respectively. 
Experimental results obtained confirm ex- 
actly these expectations. 

Gynanders also occur in moths. Whereas 
male moths usually have large, beautifully 
colored wings and females, small stumps of 
wings, gynanders have been found with 
wings like the male on one side and those 
like the female on the other side. The ex- 
planation for these exceptions is similar to 
that given for Drosophila. In the case of 
the moth, however, the gynander usually 
starts as a male zygote (XX). 

Although most gynanders in Drosophila 
and other insects in which the male has the 
heteromorphic sex chromosomes, can be ex- 
plained in this manner, some gynanders 
originate another way. In extremely rare 
cases, an abnormal egg is produced after 
meiosis which contains not one but two hap- 
loid gametic nuclei. Because polyspermy 
sometimes occurs in insects — that is, more 
than the one sperm normally involved in 
fertilization enters an egg — one of the two 
haploid egg nuclei may be fertilized by an 
X-carrying sperm, the other by a Y-carrying 
one. The resultant individual is approxi- 
mately a half-and-half gynander. This type 
of gynander can be identified if the two 
paternal (or the two maternal) haploid 
gametic nuclei are marked differently for a 
pair of autosomal genes. 

Man and Mouse 

In human beings sexual type is determined 
at fertilization, XY zygotes becoming males; 
XX zygotes, females. In early development, 
all sex organs or gonads are neutral; that is, 
they give no macroscopic indication whether 
they will later form testes or ovaries. The 
early gonad has two regions, an outer 



one, the cortex, and an inner one, the 
medulla. As development proceeds, the 
cortex degenerates in those individuals that 
carry a Y (male), and the medulla forms 
a testis; in individuals genetically determined 
to be females, the medulla degenerates, and 
the cortex forms an ovary. 

Once the testis and ovary are formed, they 
take over the regulation of further sexual 
differentiation by means of the hormones 
they produce. The hormones direct the de- 
velopment or degeneration of various sexual 
ducts, the formation of genitalia, and other 
secondary sexual characteristics. Since sex- 
ual differentiation is largely controlled by 
the sex hormones, it is not surprising that 
genetically normal individuals are morpho- 
logically variable with regard to sex. Any 
change in the environment that can upset 
the production of, or tissue response to, sex 
hormones can produce effects which modify 
the sex phenotype. So, the phenotypes nor- 
mally considered male and female show some 
variability — providing some of the spice of 
life. Genetically normal persons exposed to 
abnormal environmental conditions can dif- 
ferentiate phenotypes that lie between the 
two normal ranges of sex type, and, there- 
fore, are intersexual in appearance. Though 
it is sometimes easy to classify an individual 
as being an intersex because the person is 
clearly between the two sex norms, other 
individuals at the extremes of normality can- 
not readily be labeled normal, or intersex, 
or supersex. Intersexual phenotypes due to 
environmental factors can result either from 
genotypic males who have developed par- 
tially in the direction of female, or from 
genotypic females partially differentiated in 
the direction of male. 

Otherwise-diploid individuals are known 
who have various numbers of sex chromo- 
somes. Only one type has a single sex chro- 
mosome; this is the X0 individual, who is 
female. The typical phenotypic effect of 
this condition is called Turner's syndrome 



108 



CHAPTER 8 



(after its discoverer) and is characterized by 
the failure to mature as a woman. Turner 
type females usually do not develop breasts. 
ovulate, or menstruate. Because of vari- 
ability in the genotypic details and in the 
environment (including medical treatment), 
considerable variation occurs in the pheno- 
ls pic consequences of the X0 condition. In 
fact, one woman of this constitution is known 
to have given birth to a normal (XY) son. 
The X0 mouse is apparently less variable 
phenotypically since it always seems to pro- 
duce a fertile female. The other single sex 
chromosome type, YO. presumably lethal in 
man, is known to be lethal in mouse. 

Otherwise-diploid individuals having three 
sex chromosomes are of three types: XXX 
is female (sometimes mentally defective); 
XYY is male; XXY is male. The XXY in- 
dividual who is characteristically sterile, may 
have undersized sex organs, and may de- 
velop various secondary sexual characteris- 
tics of females, possesses Klinefelter's syn- 
drome (named after its discoverer). Along 
with the X0 female, he is phenotypically 
variable; for instance, some Klinefelter males 
are mentally retarded, others are not; al- 
though all those presently known are sterile, 
some show normal sexual drive and be- 
havior. In the mouse, XXY is a sterile 
male. 

Otherwise-diploid persons of the follow- 
ing additional types are also known: XXXX 
( 9 ) ; XXXY ( $ ) ; XXYY ( $ ) ; XXXXX 
( 9 ) ; XXXXY ( $ ); XXXYY ( $ ). Con- 
trary to the situation in Drosophila, it is 
clear from all these results that the Y chro- 
mosome is the primary sex-determining 
chromosome in man and mouse. Presence 
of a single Y determines the sex as male; 
absence of a Y produces the female. All 
individuals require an X in order to be 
viable. 

The Y versus no Y sex-determining mech- 
anism in human beings and mice implies 
that the Y must carry one or more genes 



for maleness in that portion which makes 
it cytologicaUy unique, the X having no cor- 
responding allele(s). Admitting that the 
presence of gene(s) for maleness on the Y 
makes for male, what is genetically responsi- 
ble for the femaleness produced in the ab- 
sence of the Y? Clearly other genetic factors 
are present — not limited in location to the 
Y chromosome — which affect sex and, there- 
fore, femaleness. The female tendency often 
shown by the human XXY suggests that the 
X contains genes affecting normal sexual 
differentiation which, when present in excess, 
cause a shift toward femaleness. Presum- 
ably, the X also has this capacity when Y 
is absent. 

All cases in which the entire body seems 
to contain an abnormal number of sex chro- 
mosomes can be explained as the result of 
nondisjunction leading to chromosome loss 
or gain which occurs either during meiosis 
or at an early cleavage division — probably 
the first — of the fertilized egg. Such nondis- 
junctions are correlated in human beings 
with the mother's advanced age at the time 
of pregnancy. 

By following the distribution of X-l inked 
mutants, it has been shown, however, that 
the nondisjunction which produces an ab- 
normal sex-chromosome number sometimes 
involves the paternally contributed sex-chro- 
mosome material. This origin is exemplified 
by a red-green colorblind father having an 
X0 daughter of normal vision. Since cer- 
tain aged Drosophila eggs cause the loss of 
paternal chromosomes after fertilization, it 
is important to recognize the possibility that 
the loss of a paternal chromosome in man 
can occur post- as well as pre-meiotically. 
Due to a premeiotic paternal nondisjunction 
colorblind women can, of course, have XXY 
Klinefelter sons of normal vision. 

A considerable number of persons having 
different chromosomal compositions in dif- 
ferent body parts are mosaic for sex chro- 
mosomes. These include the following 



Sex Determination 



109 



mixed constitutions: XXX/XO; XX/XO; 
XY/XO, XXY/XX; XXXY/XY. Such 
cases are usually due to one or more errors 
in chromosome distribution among the 
daughter nuclei produced after fertilization. 
Although such individuals are sex-chromo- 
some mosaics, and some may even have 
one ovary-like and one testis-like gonad, 
they are not gynanders in superficial char- 
acteristics because of their whole-body dis- 
tribution of sex hormones. Although the 
XXY male is often clearly an intersex, the 
X0, XXX, and so on females that show in- 
complete maturity are best considered infra- 
females, being underdeveloped sexually. It 
should now be clear that some specific 
phenotypic sexual abnormalities may be 
based primarily either on an abnormal en- 
vironment or on an abnormal chromosomal 
composition (recognizing also the possibility 
that mutants other than those involving an 
abnormal number of sex chromosomes can 
affect sex). Accordingly, chromosomal 
counts are often desirable in order to deter- 
mine the cause — and, hence, the treatment 
— of sexual abnormality. 

Human Sex Ratio 

Consider how the genotype is related to the 
sex ratio, that is, to the relative numbers of 
males and females born. On the average, 
106 boys are born for each 100 girls. This 
statistic might be surprising at first, since 
half the sperm are expected to carry X, half 
Y, and all eggs, an X, the ratio of boy to 
girl expected at conception is one to one. 
Even if the four meiotic products of a given 
cell in spermatogenesis usually carry X, X, 
Y, Y, there is the possibility that during or 
after spermiogenesis (conversion of the telo- 
phase II cell into a sperm) some X-bearing 
sperm are lost. This possibility is supported 
by a report 4 that human ejaculates contain 
sperm heads of two sizes and shapes (Fig- 
ure 8-6); the smaller type sufficiently in 

4 By L. B. Shettles (1960). 




figure 8-6. Head shapes in human sperm. 
Round-headed sperm are reported to be smaller 
and more numerous than oval-headed sperm, 
suggesting these carry the Y and X chromo- 
somes, respectively. {Courtesy of L. B. 
Shettles.) 



excess to explain an excess of males at fer- 
tilization provided the smaller sperm con- 
tains the small Y chromosome, and the larger 
sperm carries the larger X chromosome. 
Other evidence suggests that at conception 
males are much more numerous than fe- 
males; since more male fetuses normally 
abort than female, the numbers of boys and 
girls are more nearly equal at the time of 
birth than they were at conception. 

A study of the sex ratio at birth shows 
that the ratio 1.067:1.000 is found only 
among young parents, and that it decreases 
steadily until it is about 1.036: 1.000 among 
the children of older parents. How may 
this significant decrease be explained? Per- 
haps in older mothers there is a greater 
chance for chromosomally normal male 



10 



CHAPTER 8 



babies to abort, or tor chromosome loss in 
the earliest mitotic divisions o( the fertilized 
egg. If the chromosome lost is an X and 
the zygote is \Y. the loss is expected to be 
lethal, so that a potential boy is aborted. 
If the zygote losing an X is XX. a girl can 
still be born. Moreover, if the chromosome 
lost in the XY individual is a Y, a girl can 
be born instead of a boy. Part of the effect 
must be due to the increase in meiotic non- 
disjunction with maternal age (zygotes of 
XXX type form viable females, whereas 
zygotes of Y0 type are expected to abort). 

We must include the possibility that the 
fathers may also contribute to this shift in 
sex ratio. Postmciotic selection against Y- 
carrying sperm may increase with paternal 
age. Or. as fathers become older, the XY 
tetrad may be more likely to undergo non- 
disjunction to produce sperm containing re- 
spectively, X, X, YY, 0. The first two can 
produce normal daughters; the last one can 
produce an underdeveloped X0 daughter; 
and only the YY is capable of producing 
males. Even though the XYY individual 
is male, it may frequently abort. Other 
genetic and nongenetic explanations for the 
shift in sex ratio with age are also possible. 
This discussion merely demonstrates how 
the basic facts of sex determination, chro- 
mosome loss, and nondisjunction may be 
used to formulate various hypotheses whose 
validity is subject to test. 

When many pedigrees are examined for 
sex ratio, several consecutive births of the 
same sex occasionally occur. This phe- 
nomenon could, of course, happen purely 
as a matter of chance when enough pedigrees 
are scored. One family, however, is re- 
ported to have only boys in 47 births and, 
in another well-substantiated case, out of 
72 births in one family, all were girls. In 
both these cases the results are too improb- 
able to be attributed to chance. 

We do not know the basis for such results 
in man, but two different cases of almost 



exclusive female progeny production in Dro- 
sophila might suggest an explanation for 
those human pedigrees in which only one 
sex occurs in the progeny. In the first case, 
an XY male carrying a gene called sex ratio 
is responsible. Because of this gene, the 
X and Y fail to synapse, and the X replicates 
an extra time to form a tetrad; since almost 
all Y chromosomes degenerate during meio- 
sis, almost all sperm carry an X. In the 
second case, a female transmitting a spiro- 
chaete microorganism to her offspring 
through the egg is responsible. Such a fe- 
male mated to a normal male produces 
zygotes which begin development; soon 
thereafter the XY individuals are killed by 
the spirochaete, leaving almost all female 
survivors. 

The sex ratio can be controlled if the 
genotypes of the zygotes formed can be con- 
trolled. Since X- and Y-bearing sperm of 
men apparently differ in cytological appear- 
ance (Figure 8-8), it should be possible to 
separate them and thereby control the sex 
of progeny. Using various animal forms, 
such experiments have been performed with 
some success by Russian, American, and 
Swedish workers, using electric currents or 
centrifugation. Although these experiments 
have been encouraging, the results are not 
yet consistent, and the techniques not yet 
suitable for practical use. 

Hymenoptera 

In Hymenoptera (for example, bees, ants, 
wasps, and saw flies) unfertilized eggs de- 
velop as males (haploids) and fertilized 
eggs, usually, as females (diploids). Hap- 
loid males produce haploid sperm via suit- 
able modifications of the meiotic process, 
and all gametes of males and females have 
morphologically identical chromosomal com- 
positions. 

In the parasitic wasp, Habrobracon jug- 
landis, when the parents are closely related, 
some of the sons are haploid, but others are 



Sex Determination 



111 



diploid having ten pairs of chromosomes 
like their sisters. Genetic study shows that 
such diploid males have a biparental origin. 
Not only are diploid males relatively in- 
viable, but the hatchability of sibling eggs 
is very poor. A study of intrastrain and 
interstrain breeding supports the interpreta- 
tion 5 that a multiple allelic series determines 
sex in this form. With respect to this sex- 
determining locus or chromosome region, 
haploids are males, diploid heterozygotes 
are females, and diploid homozygotes are 
semisterile males. 

Role of Genotype in Sex Determination 

In certain organisms, male and female gam- 
etes are produced in the same individual. 
Animals of this type are said to be hermaph- 
roditic (after Hermes and Aphrodite), and 
plants, monoecious. The hermaphrodite 
snail, Helix, has a gonad which produces 
both eggs and sperm from cells which some- 
times lie very close together. In the earth- 
worm, eggs and sperm are produced in sepa- 
rate gonads located in different segments of 
the body. In certain mosses, egg and sperm- 
like gametes are also produced in separate 
sex organs (located on the same haploid 
gametophyte). 

In all these cases, the two types of gametes 
are produced by an organism that has but 
a single genotype; that is, one that is not 
genetically mosaic. Nevertheless, it might 
be supposed, at first, that the haploid geno- 
type carried by eggs and by the sperm is 
different and causes the difference in pheno- 
type and behavior. In the case of the 
gametophyte of mosses, however, the indi- 
vidual is haploid and so are both types of 
gametes it forms. Accordingly, in such or- 
ganisms we cannot expect differences in 
gene content to be the basis either for the 
formation of gametes or for the different 
types of gametes produced. 

5 See P. W. Whiting (1943). 



Gamete formation in hermaphroditic and 
monoecious organisms, therefore, must de- 
pend primarily upon environmental differ- 
ences. Such differences must exist even 
between cells which lie close together, as is 
the case in Helix. It is reasonable to sup- 
pose that the same kinds of environmental 
factors which can direct one group of cells 
to form muscle cells and an adjacent group 
to form bone cells, can direct the differen- 
tiation of still other cells to make gonadal 
tissue in which adjacent cells can further 
differentiate as sperm and egg. 

Note, however, that sex involves another 
kind of differentiation, which, at least in 
organisms like the mosses, is separate from 
the type of gamete formed. This problem 
(which will not be discussed in detail here) 
concerns the genetic and environmental fac- 
tors responsible for the onset of meiosis, 
which is, of course, the feature most funda- 
mental to the success of the sexual process 
as it presently occurs in many species. 

In the examples already mentioned, the 
type of gamete differentiated depends upon 
the different positions which cells have 
within a single organism; consequently, they 
are subject to differences in internal and 
external environments. In the marine an- 
nelid, Ophryotrocha, the two sexes are in 
separate individuals, and the sex type formed 
is determined by the size of the organism. 
When the animal is small, because of youth 
or because it was obtained by amputation 
from a larger organism, it manufactures 
sperm; when larger, the same individual 
shifts to the manufacture of eggs. In this 
case the environment of the gonad is changed 
by the growth of the organism. 

Finally, consider sex determination in the 
marine worm, Bonellia, in which the sepa- 
rate sexes are radically different in appear- 
ance and activity — females being walnut- 
sized and having a long proboscis, males 
being microscopic ciliated forms that live 
as parasites in the body of the female. Fer- 



112 



CHAPTER 8 



tilized eggs, grown in the absence of adult 
females develop as females; they develop 
as males in the presence either of adult fe- 
males or simply an extract of the female's 
proboscis. In this case, then, differentiation 
as a whole including sexual differentiation, 
is regulated by the presence or absence of a 
chemical messenger manufactured by fe- 
males. 

Nothing has been stated about the specific 
genetic basis for the determination or dif- 
ferentiation of sex in any of the examples 
given in this section because different sexes 
or gametes are determined not by genetic 
differences between cells, organs, or indi- 
viduals, but by environmental differences 
acting upon a uniform genotype. The genes, 
nevertheless, must play a role in all these 
cases by making possible different sexual 
responses to variations in the environment. 

Importance of Sexuality 

Even if reproduction occurred only by asex- 
ual means, the earth would now be popu- 
lated by genetically different kinds of or- 
ganisms, each variant having arisen by muta- 



tion in a pre-existing individual who was, 
in turn, produced from an unbroken line 
of descent. This method of direct descent 
is inefficient, however, since biologically fit 
individuals must wait for the rare occurrence 
of mutation to make them more fit. 

The biological innovation of sexuality has 
a tremendous genetic advantage over ascx- 
uality by providing genetic recombination 
which speeds up the process of the evolu- 
tion of more adaptive organisms. A more 
adaptive genotype may be produced in one 
individual by the combination of allelic and 
nonallelic genes originally located in two 
parents who, individually, may have been 
less well or even poorly adapted. Since 
genetic recombination normally occurs each 
generation for each gene pair, adaptive com- 
binations of genes originate much more rap- 
idly by recombination than by the relatively 
rare event of mutation. It should be clear, 
therefore, that sexuality, which produces a 
greater variety of adaptive genotypes in a 
given period of time than asexuality, is pri- 
marily responsible for the great variety of 
adapted kinds of individuals that have ap- 
peared on the Earth in recent times. 



SUMMARY AND CONCLUSIONS 

An understanding of the basis of sex requires the answer to two questions: What is 
responsible for the onset of meiosis? What is the basis for the formation of different 
kinds of gametes? Only the latter question is discussed in significant detail. In some 
cases the environment and in other cases the genotype is primarily responsible for sex 
determination. In the latter cases, sexual differences can often be correlated with 
cytogenetic differences. 

Genes responsible for sex determination are located not only in the sex chromosomes 
but in the autosomes as well. Although sex type may be changed through the action 
of a single pair of genes, a given sex is usually the result of the interaction of several, 
and probahly many, pairs of genes. Sex is, therefore, a polygenic trait (Chapter 5). 

Chromosomal differences found among zygotes serve as visible manifestations of 
differences in the balance of genes concerned with sex. Whenever, as in female 
Drosophila, genie balance is unaffected by the addition or subtraction of whole sets of 
chromosomes, sex also is unaffected. However, changes in chromosome number which 
produce intermediate genie balances also produce intermediate sex types — intersexes; 
those which make the balance more extreme than normal produce extreme sex types — 
supersexes. 



Sex Determination 113 

These principles of sex determination apply also to human beings. In man and 
many other organisms, a large part of sexual differentiation is controlled by sex hor- 
mones produced by the gonads. This type of control rarely, if ever, permits the occur- 
rence of individuals who are typically male in one part and typically female in another 
part; it may also contribute to the formation of abnormal sex types for nongenetic 
reasons. 



REFERENCES 

Bangham, A. D., "Electrophoretic Characteristics of Ram and Rabbit Spermatozoa," 
Proc. Roy. Soc, Ser. B, 155:292-305, 1961. 

Bridges, C. B., "Sex in Relation to Chromosomes and Genes," Amer. Nat., 59:127-137, 
1925. Reprinted in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood 
Cliffs, N.J.: Prentice-Hall, 1959, pp. 117-123. 

Goldschmidt, R. B., Theoretical Genetics, Berkeley and Los Angeles: University of 
California Press, 1955. 

Hannah-Alava. A.. '"Genetic Mosaics," Scient. Amer., 202:118-130, 1960. 

Lancet, No. 7075, Vol. 1, 1959, pp. 709-716. 

McKusick, V. A., Human Genetics, Englewood Cliffs, N.J.: Prentice-Hall, Inc., 1964. 

Shettles, L. B., "Nuclear Morphology of Human Spermatozoa," Nature, London, 186: 
648-649, 1960. 

Shettles, L. B., "Nuclear Structure of Human Spermatozoa," Nature, London, 188- 
918-919, 1960. 

Sturtevant, A. H., "A Gene in Drosophila melanogaster that Transforms Females into 
Males," Genetics, 30:297-299, 1945. 

Whiting, P. W., "Multiple Alleles in Complementary Sex Determination in Habro- 
bracon," Genetics, 28:365-382, 1943. 



QUESTIONS FOR DISCUSSION 

8.1. If sexual reproduction is as advantageous as discussed, why do so many organ- 
isms still reproduce asexually? 

8.2. Does the study of sex determination offer any test of the theory that chromo- 
somes furnish the physical basis for genes? Explain. 

8.3. Is it possible to consider the factors responsible for the meiotic process separately 
from the factors responsible for gamete formation? Explain. 

8.4. Why is meiosis the most fundamental feature in the success of sexuality? 

8.5. Give the genotypes and phenotypes of the unexceptional, the nondisjunctional, 
and the gynandromorphic offspring expected from a mating of f 34b /f with / 
Drosophila. 

8.6. Are there isoalleles for the genes determining the size and shape of the bristles 
and hairs of Drosophila? Explain. 

8.7. Using first the autosomal alleles e and e+ and then the X-linked alleles y and y + , 
devise crosses by which you could identify gynanders in Drosophila resulting 
from two fertilizations of a single egg. 



Ill ( HAPTER 8 

S.S. Compare the genotypes and phenotypes of sex chromosome mosaics of flies, 
mollis, and men. 

8.9. All human beings have the same number o\ chromosomes in each somatic cell. 
Discuss this statement giving evidence in support of your view. 

8.10. The following types oi mosaics are known in human beings: 

XXX/XO XXY/XX 

XX/XO XXXY/XY 

XY/XO 

Give a reasonable explanation lor the probable origination of each. 

8.11. In human beings, can the members of a pair of monozygotic twins ever be of 
different sexes? Explain. 

S.I 2. Does a gene have to have an alternative allele before it can be discovered? 
Explain. 

8.13. Assuming each homolog carried a different allele, a 1 , a 2 , a'\ o\ the same gene, 
make a schematic representation of a trivalent as it might appear during synapsis. 
Show diagrammatically the chromosomal and genetic content of the four meiotic 
products that could be obtained from your trivalent diagram. 

5.14. Ignoring chiasma formation, how many chromosomally-different kinds of eggs 
can a triploid Drosophila female produce? How many of these eggs have more 
than a 5% chance of occurring? 

8.15. (a) Scurfy, sf, is an X-linked recessive gene that kills male mice before they repro- 
duce. How is a stock containing this gene maintained normally? 

(b) Occasionally, the stock containing this gene produces scurfy females which 
also die before reproductive age. Suggest a genetic explanation for these female 
exceptions. Describe how you would test your hypothesis genetically by trans- 
planting ovaries and obtaining progeny from them. 

8.16. What explanations can you offer, other than those already presented, for the 
shift in sex ratio with age of human parents? 

8.17. No Y0 human beings are known. Why is this chromosomal constitution con- 
sidered to be lethal? 

8.18. List the types of human zygotes formed after maternal nondisjunction of the 
X chromosome. What phenotypes would be expected for each of the zygotes 
that these, in turn, may produce? 

8.19. List specific causes for the production of abnormal sex types in human beings. 

8.20. How can you explain that only one X0 individual is known to have had a suc- 
cessful pregnancy, whereas other XO's are sterile? 

8.21. Discuss the general applicability of the chromosomal balance theory of sex 
determination. 

8.22. In Drosophila, why are gynanders not intersexes? Is this true in man also? 
Explain. 

8.23. What chromosomal constitution can you give for a triploid human embryo that 
is "male"? ■"Female"? 

8.24. A non-hemophilic man and woman have a hemophilic son with Klinefelter's 
syndrome. Describe the chromosomal content and genotypes of all three indi- 
viduals mentioned. 



Sex Determination 115 

8.25. A white cat reported by H. C. Thurline, had one yellow and one blue eye, a 
phallus and one testis, one uterine horn, and one ovary. Although the animal 
had 38 chromosomes (the normal diploid number), some nuclei had an XX 
and others, an XY content. Suggest hypotheses to explain the chromosomal 
content of this individual. 

8.26. Give a possible chromosomal formula for human individuals who are: 

(a) Triploid males 

(b) Klinefelter's type of male 

8.27. A 26-year-old somewhat mentally retarded man is known to be XYY but other- 
wise diploid. To what do you attribute the apparent rarity of this type of 
chromosomal constitution? 

8.28. Klinefelter-type males occur who are XXXYY. Give a possible origin of this 
chromosomal constitution. 

8.29. In the insect Protenor and certain short-horned grasshoppers, all eggs have the 
same number of chromosomes, and half of the sperm are different, in that they 
carry one less chromosome. What is the cytogenetic basis for sex determination 
in such cases? 

8.30. In the plant genus, Melandrium, one observes individuals of the following types: 
Diploid: XX + 11AA-5 XY + 1 1 AA = 6 

Triploid: XXX + 1 1 AAA = 9 XXY + 1 1 AAA = <$ 

Tetraploid: XXXX + 1 1 A AAA = $ XX YY + 1 1 AAAA = $ 

or 
XXXY + 1 1 AAAA = $ 
Discuss the cytogenetic basis for sex determination in Melandrium. 

8.31. Discuss the sex ratio expected in the honey bee from unmated and mated females. 

8.32. Does the chromosomal balance hypothesis of sex determination apply in the 
case of parasitic wasps? Explain. 

8.33. Compare the self-sterility alleles in Nicotiana (see p. 60) with the sex-determination 
alleles in Habrobracon. 



Chapter 9 

LINKAGE AND CROSSING OVER 
BETWEEN GENES 



T: 



Jhe alleles of a gene pair af- 
fecting the seed coat of the 
garden pea were symbolized 
in Chapter 4 as round (/?) and wrinkled 
(r). This symbology follows the conven- 
tion that uses upper and lower case of the 
first letter (or so) of the phenotype pro- 
duced by the dominant allele — the one usu- 
ally found in nature — to represent the domi- 
nant and recessive alleles, respectively. 

In other conventions (see Figure 9-1), 
the first letter (or so) of the recessive trait 
(wrinkled) is used in lower case for the 
recessive allele (w), and the normally dom- 
inant allele (round) is given as one of the 
following: the same symbol in upper case 
(W); a + symbol as a superscript or base 
to the lower case symbol (vv+ or +"); or 
-+- alone. Henceforth in this book we will 
usually use one form of the -f- system for 
symbolizing genes. In this system, a mutant 
gene — Beadex, for example — which is dom- 
inant to the normal wild-type allele is rep- 
resented by one (or more) letters of which 
the first is capitalized (Bx or + "■'') and its 
wild-type allele is + (or Bx+). The hy- 
brid + vv can be represented as = or or 
r WW 

-\-/w to show that these alleles are on dif- 
ferent members of a pair of homologous 
chromosomes. 

Each of the first seven pairs of genes 

studied in the garden pea (see Chapter 4, 

p. 48) appeared to segregate independently. 

If this kind of segregation is attributed to 

116 



each gene pair being located in a dilferent 
one of the seven pairs of chromosomes car- 
ried by this organism, what result will be 
obtained when an eighth pair of genes, 
showing dominance and affecting an unre- 
lated trait, is included in such a study? 
When a dihybrid is made of one of the 
seven gene pairs and the eighth pair men- 
tioned above, a 9:3:3:1 phenotypic ratio 
is obtained when the dihybrid is self-ferti- 
lized, and a 1:1:1:1 phenotypic ratio is 
obtained when the same dihybrid is test 
crossed to the double recessive. These two 
independent tests demonstrate that the two 
pairs of genes involved are segregating in- 
dependently. The phenotypic ratios are 
radically different, however, when a dihy- 
brid is made with still another of the seven 
gene pairs — the one affecting seed coat — 



W 
w 



+ w 
w + 



w 



w 



+_ 
w 



— +/ w 
w 



figure 9-1. Various ways of representing the 
round-wrinkled hybrid by gene symbols. 

and the same eighth. The other pair of 
genes involved (the eighth) determines 
the presence and absence of tendrils — the 
threadlike structures serving as a means for 
attachment as the plant climbs. The ten- 
drilless allele (t) is recessive. When a dou- 
ble recessive pea plant — wrinkled, tendril- 
less (w w 1 1) is crossed to a pure double 
dominant — round, tendrils ( + + + + ), all 
Fi are round with tendrils (+w +/), as 
expected. When the F, are self-fertilized 
(dihybrid by dihybrid), the following re- 
sults are obtained in F>: 



Phenotype 


No. 


Individuals 


round, tendrils 




319 


round, tendrilless 




4 


wrinkled, tendrils 




3 


wrinkled, tendrilless 




123 



Linkage and Crossing Over Between Genes 



117 



Note that each gene pair shows segrega- 
tion in the F 2 since the ratio of round to 
wrinkled is 323:126 (a 3:1 ratio), and the 
ratio of tendrils to no tendrils is 322:127 
(a 3:1 ratio). Had these gene pairs been 
segregating independently, the resultant ra- 
tio would have been 9:3:3:1. Instead, the 
F L » has relatively too many plants pheno- 
typically like the P x parents (wrinkled, no 
tendrils; round, tendrils) and relatively too 
few new recombinational types (round, no 
tendrils; wrinkled, tendrils). 

Examine also the phenotypic results ob- 
tained from test crossing the dihybrid in 
question ( + w -\-t by w w t t) : 



Phenotypes 


No. 


Individuals 


round, tendrils 




516 


round, tendrilless 




9 


wrinkled, tendrils 




7 


wrinkled, tendrilless 




492 



Independent segregation would have 
given a 1:1:1:1 ratio for each of the types. 
But again the dihybrids produced relatively 
excessive numbers of gametes containing 
the old (parental) combinations (-\ — h and 
w t) and relatively too few new combina- 
tional or recombinational types. Based on 
the results of both crosses, we conclude that 
independent segregation does not occur in 
this dihybrid. The very existence of re- 
combinational types proves — what had pre- 
viously been an assumption — that we are 
dealing with two separate pairs of genes. 

Let us assume now that the two pairs of 
nonalleles involved are located in the same 
pair of homologous chromosomes, a possi- 
bility already mentioned in Chapter 4 (p. 
48). In this situation the nonalleles in the 
same chromosome are linked to each other. 
Recall that sex-linkage involves the linking 
of a single gene (such as the one for white 
eye in Drosophila) to a particular chromo- 
some (the X chromosome). Our concern 
here is with intergenic linkage, which in- 



volves all the nonallelic genes presumed to 
be located in the same chromosome. We 
can obtain evidence for this only by study- 
ing the transmission genetics for at least two 
traits simultaneously. Since no genetic re- 
combination was detected between the ge- 
netic material for sex and for a sex-inde- 
pendent trait (like eye color) in the X chro- 
mosome, the linkage between the two traits, 
sex-linkage (or, more precisely in this case, 
X-linkage) was complete and presented no 
evidence that this chromosome contained 
two or more separable nonalleles. Because 
the present experiments with peas involved 
two separable pairs of genes, we were able 
to propose the hypothesis that a chromo- 
some contains more than one gene. 

Let us reexamine the results of the two 
kinds of pea crosses described. In Figures 
9-2 and 9-3 a horizontal line is used to 
represent a chromosome and to indicate the 
presence of one member of each pair of 
alleles in each chromosome. Where the 
genes could be either the dominant or the 
recessive allele, a question mark is placed 
in the appropriate position. Down through 
the genotypes of the P 2 the results in Fig- 
ure 9-2 are consistent with the view that 
linkage is complete; that is, the chromo- 
somes carrying w t or + -f are unchange- 
able (except by mutation). However, the 
occurrence of seven recombinational indi- 
viduals in F 2 shows that linkage is not com- 
plete — that these recombinants have a chro- 
mosome which has kept one allele and re- 
ceived the nonallele present in the homolog. 
Moreover, reciprocal types of recombinants 
are approximately equal in frequency, sug- 
gesting that a given pair of genes switched 
positions in the homologs; that is, they had 
reciprocally crossed over. For this reason, 
such recombinational individuals are said to 
carry a crossover chromosome produced by 
a process called crossing over. Therefore, 
complete linkage between genes is prevented 



IIS 



CHAPTER 9 



Wrinkled, no tendrils x Round, tendrils 
wt + + 



wt 



F ] Round, tendrils ^z^ 

wt 

P, F, Round, tendrils (self-fertilized) 

w t w t 

+ + 
F Round, tendrils 

Round, no tendrils 

Wrinkled, tendrils 

Wrinkled, no tendrils 



mgurf. 9-2 (above). Linkage between non- 
allelic genes in the garden pea. 



figure 9-3 (below). Linkage between non- 
allelic genes in the garden pea. The dihybrid 
parent is the same as the F, in Fig. 9-2. 

P^ F | Round, tendrils x Wrinkled, no tendrils 

+ + w t 





319 


? ? 




+ t 






4 


? t 


w + 






3 


w? 


wt 






123 


wt 


TOTAL 


449 



w t 



Round, tendrils 



Round, no tendrils 



Wrinkled, tendrils 



Wrinkled, no tendrils 





O 10 


w t 




+ t 






9 


w t 




w • 


7 




w t 




w t 


492 


w t 




TOTAL 


1024 



In a crossing-over process that produces 
genetic recombinations called crossovers. 

What other characteristics can we estab- 
lish lor the crossing-over process and the 
crossovers it produces? Among the prog- 
eny obtained from backcrossing the dihy- 
brid ( Figure 9-3 ) . 16 received crossovers 
in the gametes contributed by the dihybrid. 
1008 did not. Again, the reciprocal cross- 
over classes are about equal in frequency. So 
approximately one crossover was produced 
for each 63 noncrossovers. A simple cal- 
culation will show that the F_. results in Fig- 
ure 9-2 are consistent with this proportion. 

These genes can also make a dihybrid 
which receives one mutant (recessive) and 
one normal (dominant) gene from one par- 
ent (w+) and one normal and one mu- 
tant gene from the other parent (-M). 
When such a dihybrid is test crossed, the 
crossovers (w(or -j- +) and noncrossovers 
(w -\- or -f O also occur in the proportion 
1:63. Crossing over, apparently, occurs 
with the same frequency whether the two 
mutant genes enter the dihybrid from the 
same parent or from different parents. 
Crossovers, therefore, occur in the gametes 
of an individual with a frequency that is 
constant and independent of the specific 
combination in which the nonalleles were 
received. If this is the normal behavior, it 
must follow that even in -| — \-/-\ — |- and 
w t/w t individuals, one gamete in each 64 
produced is a crossover for these genes but 
undetected because it carries no new com- 
bination of nonalleles. Notice that the 
crossover progeny are fewer than the non- 
crossover progeny. This must mean that 
when two linked mutants enter a dihybrid 
in the same gamete, the mutants tend to be 
transmitted together to the gametes made 
by this dihybrid (coupling); if, on the other 
hand, the mutants enter the dihybrid sepa- 
rately, they tend to be transmitted separately 
to the next generation (repulsion). 



Linkage and Crossing Over Between Genes 



110 



In another species, the sweet pea, the 
trait purple flowers is due to a single gene 
( + ) whose recessive allele (r) produces 
red flowers. Long pollen ( + ) is dominant 
to round pollen (ro). Assume two pairs 
of genes are involved in a cross between a 
pure line of purple long (-)- -+-/+ +) and 
red round (r ro rro). The F] produces all 
purple long (-| — \-/rro) and self-fertiliza- 
tion of the F, produces in F 2 too many P, 
phenotypes and too few new recombina- 
tional types (purple round and red long) 
for independent segregation. Therefore, 
these genes must be linked. But, as before, 
linkage is incomplete. 

In this case, the crossovers obtained can 
be accounted for if the Pj(Fj) dihybrid 
forms gametes in the relative proportions 
10 + -f:10rro:l + ro:l r + . This fre- 
quency of crossovers is obtained no matter 
how the genes enter the dihybrid. Notice, 
however, that the constant frequency (1/11) 
in the sweet pea differs from the frequency 
(1/64) observed previously in the garden 
pea. 

Consider also, the following cases: 

1 . In Drosophila you recall, the mutant 
gene (w) for white eye is X-linked. So 
also is another (presumably nonallelic) mu- 
tant gene which produces miniature wings 
(m). Using pure lines, a white-eyed long- 
winged fly is crossed to a dull-red-eyed min- 
iature-winged fly. The Fi female carries 
two X's and is, presumably, w +/+ m - 
This female is then mated, and the sons 
are scored phenotypically. (Any male can 
be used as parent since it will usually trans- 
mit to each son a Y chromosome lacking 
alleles of the genes under consideration. In 
fact, the Y is found to lack alleles of almost 
all of the genes known to be present on the 
X except the gene for bobbed bristles, bb. 
Moreover, the Y contains several genes for 
male fertility that have no alleles on the X.) 
Since sons normally receive their single X 



from their mother, their phenotypes directly 
indicate which hemizygous X-linked alleles 
each has received. Among the sons of this 
mating, about one crossover type appears 
for every two that are noncrossovers. 

2. In man. color-blindness (c) and he- 
mophilia type A (h) are recessive X-linked 
mutant genes absent on the Y chromosome. 
Though rare, some women have the geno- 
type + h/c H with one of these mutants 

on each X. Available data indicate that 
crossover (c h or + +) and noncrossover 
( + h or c + ) sons occur in the approxi- 
mate ratio of 1:9. 

These examples show that when linkage 
between nonalleles is incomplete, the per- 
centage of progeny carrying crossovers is 
constant for a given case but can be quite 
different in different organisms. 

The possibility that the strength of link- 
age varies in the same organism can be 
tested using two mutants, b (black body 
color) and vg (vestigial wings), of Drosoph- 
ila melanogaster. 

A P, cross between vg -f- vg -f- (vestig- 
ial) 1 females and + b /+ b (black) males 
produces all normal F, (vg + /+ b). As 
shown in Figure 9-4A, a test cross of the Fj 
female {vg + + b 9 by vg b/vg b $ ) pro- 
duces in Fj only 20% with recombinant 
chromosomes. (All Fj carry vg b from the 
father, their maternal chromosome 40% of 
the time is vg +; 40%, + b; 10%, + +; 
10%, vgb.) Since these results are inde- 
pendent of sex, we conclude that b and vg 
are linked autosomally. Recall that the X- 
1 inked genes m and w showed 33% cross- 
overs; therefore, the linkages between differ- 
ent pairs of nonalleles on different pairs of 
homologs can have different strengths. 

When the reciprocal cross ( vg -f-/+ b 6 

1 The convention used here, and usually hereafter, 
is to describe the phenotypes of individuals only 
with respect to the appearance of mutant traits, 
all traits not mentioned being of the normal type. 



120 



CHAPTER 9 



r \ vg + 

vg \ 



9* Ho* 



dW °" d ?? 



vg 



+ b 



A 



_ + F o x nip^ 



vg b 



vg b 
vg b 



B 

? 



vg + _ 

x F. 

+ b ' 



a* 



F 2 40% 



40% 



10% 



10% 



vg 


+ 


vg 


b 


+ 


b 


vg 


b 


+ 


+ 


vg 


b 


vg 


b 



vg b 



F 2 50% 



50% 



vg + 
vg b 



+ b 
vg b 



figure 9-4. Results of reciprocal crosses involving black body color (b) and vestigial 
(vg) wings. 



by vg b/vg b $ ) is made with the Fi, 50% 
of offspring are vg-\-/vgb (vestigial), and 
50% are + b/vg b (black) (Figure 9-4B). 
This cross produces no offspring with cross- 
overs, so that linkage is complete for these 
genes in the male Drosophila. (Had link- 
age been complete in the female also, we 
should not have had any evidence that vg 
and b are separable and, therefore, two 
genes instead of one. ) One finds, more- 
over, that in Drosophila any genes showing 



incomplete linkage in the female are com- 
pletely linked in the male; the male, there- 
fore, does not undergo the process of cross- 
ing over to produce crossovers. 2 It may be 
noted that in animals in general, crossing 
over is reduced or absent in the hetero- 
gametic sex. For example, no crossing over 
occurs in the females of birds. 

2 On rare occasions a special kind of "crossing 
over" does occur in the male Drosophila but is 
not of the kind that typically occurs in females. 



Linkage and Crossing Over Between Genes 



121 



What is the strength of linkage between 
a given gene and several nonalleles located 
in the same chromosome? This problem 
can be readily studied for certain X-linked 
genes in Drosophila. In Figure 9-5, one 
column shows genotypes of females; the 
other column shows the frequencies of 
crossover combinations as detected in their 
sons. The recombination frequencies given 
are those found between the gene for yel- 
low body color (y) and for each of the fol- 
lowing: white eye (w); crossveinless wings 
(cv); cut wings (ct); miniature wings (m); 
forked bristles (/). For example, 13 of 
each 100 eggs produced by the female di- 
hybrid for v and cv carry crossovers (-| — f- 
or ycv). What does this value, and the 
other still different linkage values, mean in 
terms of meiosis? 

So far, no commitment has been made 
as to where or when crossing over takes 
place. Since we have been concerned with 
complete and incomplete linkage as studied 
in successive generations of individuals, let 
us consider only crossing over that occurs 
in the cell line that gives rise directly to the 



FEMALES 


% CROSSOVER 
CHROMOSOMES 
AMONG SONS 


y + / + w 




1.5 


y + / + cv 




13 


y + / + ct 




20 


y + / + m 




34 


Y + / + I 




48 





figure 9-5. Crossover percentages between 
one gene and others linked to it. 



figure 9-6. The genetic consequences ex- 
pected after a crossing over between linked 
genes. 



gametes (the germ line), ignoring the pos- 
sibility that crossing over occurs in somatic 
(nongerminal) cells. Although crossing 
over might be premeiotic, meiotic, or post- 
meiotic in occurrence, we shall assume that 
all crossovers are produced during meiosis. 
The genetic consequences of an exchange 
(which we now call crossing over) between 
two pairs of linked genes during meiosis 
were discussed earlier (Chapter 4, p. 48) 
and the assumption was made (pp. 19-22) 
that a chiasma represents physical cytolog- 
ical evidence that a crossing over has oc- 
curred. 

These cytogenetic events are diagramed in 
Figure 9-6 in somewhat more detail than 
those originally shown (Figure 4-8). In 
stage I, one member of a pair of homolo- 
gous chromosomes (hollow bar) is carrying 
the recessives a and b and the other (solid 
bar) is carrying their normal alleles. The 
black dots represent centromeres. The 
homologs synapse and form a tetrad (each 



( II \I'I ER 9 



univalent is now represented b\ two sister 
•strands). After crossing over, the tetrad 
seems to appear at diplonema as depicted in 

stage 11. which shows a chiasma between 
the a and b loci (the places in a chromo- 
some containing the genes). Note that 
when the univalents are initially identical 
in appearance, a chiasma shows there was 
a physical exchange of apparently exactly 
equivalent segments between two nonsister 
strands oi a tetrad, the strands being just as 
long after as before the exchange. Stage 
111 shows the dyads present after the first 
meiotic division is completed. The upper 
cell or nucleus contains one -f + noncross- 
over strand and one + b crossover strand, 
whereas the lower one contains the recip- 
rocal crossover strand a + and the non- 
crossover strand a b. Stage IV shows the 
four haploid products (cells or nuclei) pro- 
duced after the dyads form monads, and the 
second meiotic division is completed. Ac- 
cording to this hypothesis, if one chiasma 
(representing a crossing over) occurs in any 
position between the loci of a and b, two 
of the four haploid nuclei produced con- 
tain noncrossover parental combinations, 
and the other two contain crossover non- 
parental recombinations. 

Evidence that the crossovers found in 
gametes originate in this way is ordinarily 
difficult to obtain because, in females, only 
one of the four haploid products from each 
nucleus entering the meiotic divisions is 
usually retained as the nucleus of a func- 
tional gamete, the others being lost (as 
polar body nuclei or cells). Even when 
each of the four haploid products becomes 
or gives rise to a gamete, as in sperm or 
pollen formation, the four gametes — pro- 
duced from a cell containing a given chi- 
asma — mix with gametes produced from 
other meiotic cells which may or may not 
have had a similar chiasma. For these rea- 
sons, only one of the four meiotic products 



is normally observed or identified at a time. 
If each chiasma results from a prior cross- 
ing over in the four-strand stage, approxi- 
mate^ equal numbers o\' the two reciprocal 
kinds o\ crossovers would be expected, as 
seen in the crossover data already pre- 
sented. However, crossing over during the 
two-stranded stage 1 is also expected to pro- 
duce this result. The occurrence of non- 
crossover types, which are equally frequent 
and more numerous than the crossovers, 
can be explained if crossing over between 
the loci of a and b occurs less than 509? 
of the time at the two-strand stage or less 
than 100% of the time at the four-strand 
stage. The morphology of a chiasma, how- 
ever, supports the view that crossing over 
takes place sometimes, if not always, at the 
four-strand stage. 

Genetic evidence as to whether crossing 
over occurs at the two-strand or the four- 
strand stage might be obtained from gam- 
etes that retain not one but two or more 
strands of a tetrad. Finding a gamete that 
carries one strand which is a noncrossover 
and another homologous one which is a 
crossover, would support only the four- 
strand hypothesis. A suitable system for 
this test is found in Drosophila females 
whose two X's are not free to segregate 
since they are joined and have a single cen- 
tromere. One type of such attached-X's is 
V-shaped at anaphase. During meiosis this 
attached-X replicates once, and the four 
arms synapse to form a tetrad, yielding two 
meiotic products each of which carries at- 
tached-X's and two products devoid of X 
chromosomes. Using females whose at- 
tached-X's are dihybrid and scoring their 
female progeny, one finds attached-X's hav- 
ing one arm a crossover and one that is not 
(Figure 9-7). Though this evidence also 
supports the four-strand hypothesis, it does 
not eliminate crossing over at the two-strand 
stage. 



Linkage and Crossing Over Between Genes 



123 



METAPHASE I 



w 


f 


w 


-» 
\ 
% 

f 


w a 


f 34b)| 


w a 


f 34b / 
— / 



B 



w 


f 


w 


Vi '' 


w a /\f34b; 
w a f 34b^ 



w 


f 


w a 


f 34bi 




w 


f 




\ 



TELOPHASE II 



light apricot 
weak forked 



light apricot 
weak forked 



34b 




white 
weak forked 



apricot 
weak forked 



figure 9-7. Genotypic and phenotypic consequences of no crossing over {A ) and 
of one type of crossing over (B) between marker genes in an attached-X female of 
Drosophila. 



124 



CHAPTER 9 



The red bread mold Neurospora ma) 
provide critical evidence as to the time of 
crossing over. Recall that in the sexual 
process, so-called "fruiting" bodies arc 
formed composed of cells containing two 
haploid nuclei, each of which was derived 
originall) from a different parent (Figure 
9 8). Two such haploid nuclei fuse to 
form a diploid nucleus containing seven 
pairs of chromosomes, and the cell elon- 
gates to form a sac. The diploid nucleus 
immediately undergoes meiosis, as shown in 
the figure, so that the four haploid products 
are arranged in tandem; that is. the two 
top nuclei come from one first-division nu- 
cleus, the bottom two from the other first- 



Q b 



X 




DIPLOID NUCLEUS 



DIPLONEMA 



D AFTER 

FIRST DIVISION 



FOUR 

MEIOTIC 

PRODUCTS 



JO I TWO HAPLOID NUCLEI 



DIPLOID NUCLEUS 



DIPLOID NUCLEUS 



FIRST 

MEIOTIC DIVISION. 




SECOND 

MEIOTIC DIVISION 



MITOTIC DIVISION AND 
SPORE FORMATION 




EIGHT SPORES 



figure 9-9. Chiasma and crossing over in 

Neurospora. 



figure 9-8. Meiosis in Neurospora. 



division nucleus. Since each haploid nu- 
cleus subsequently divides mitotically once, 
each meiotic product is present in duplicate 
within the ascus. Each haploid ascospore 
can be removed from the ascus, grown in- 
dividually, and its genotype determined di- 
rectly. In this organism, then, all of the 
meiotic products derived from a single dip- 
loid nucleus can be obtained and identified. 
Using the symbols in Figure 9-6, let us 



Linkage and Crossing Over Between Genes 



125 



follow in Figure 9-9 the genetic conse- 
quences of a single crossing over between 
the loci under study. Since only one of 
the seven pairs of chromosomes present is 
being traced, the others were omitted from 
the figure. As shown, a single crossing over 
in the four-strand stage produces two cross- 
over and two noncrossover meiotic prod- 
ucts. On the other hand, a crossing over 
in the two-strand stage (in the topmost nu- 
cleus) would produce only crossover meio- 
tic products. 

When numerous asci of a particular di- 
hybrid for linked genes were tested, 90% 
had all eight spores noncrossovers for the 
two loci; in the remaining 10%, exactly 
four of their eight ascospores were cross- 
overs. In other words, never were all eight 
spores from a single sac crossovers. This 
fact demonstrates conclusively that crossing 
over occurs only in the four-strand stage, 
as depicted in Figure 9-9 and Figure 9-10. 

It has already been implied that chiasma 
formation is a normal part of meiosis (p. 
16). The chiasma prevents the premature 
separation of dyads by holding them to- 
gether as a tetrad until anaphase I. (Usu- 
ally at least one, and as many as six chi- 
asmata occur per tetrad.) Therefore, the 
crossing over that subsequently leads to the 
useful chiasma is also a normal part of mei- 
osis. 

Since chiasmata are found at numerous 
positions along a chromosome, it seems rea- 
sonable to suggest that the greater the dis- 
tance between two loci, the greater will be 
the chance for a crossing over to occur be- 
tween them, and the greater will be the 
frequency of crossovers for them. Con- 
versely, the closer two loci are, the smaller 
will be the chance that crossing over oc- 
curs between them, and the smaller will be 
the frequency of crossovers for them. Ac- 
cording to this view, the frequency of cross- 
overs can be used as an indication of the 



relative distances between loci. (The re- 
sults presented in Figure 9-5 should now 
have additional meaning for us.) 

In the particular Neurospora test men- 
tioned, no crossing over occurred in 90% 
of spore sacs in the genetically marked re- 
gion (a-b). These sacs produced 90% of 
the total number of spores and carried only 
parental, noncrossover genotypes. From 
the 10% of spore sacs which did undergo 




TETRAD 





FIRST 

MEIOTIC 

DIVISION 








) SECOND 

MEIOTIC 

"\ DIVISION 


v m 






+ 




th ^ 












th 




+ 










-•- 


th 


(-•- 


th 



8 Spores 



8 Spores 



+ 




th 


+ 
th 


MITOTIC 


th 




DIVISION 




th 




+ 


th 




th 


th 




th 



figure 9-10. Arrangement of spores in the 
Neurospora ascus when segregation occurs at 
the first meiotic division {left) and at the 
second meiotic division {right), as determined 
by the absence and presence, respectively, of a 
chiasma between the segregating genes and the 
centromere. 



L26 



CHAPTER 9 



crossing over, half oi the spores were of 
the parental types and half were crossovers. 
So. equating the chiasma with the crossing 
over, a chiasma Frequency of 109? resulted 
in 5 r 't of all spores having crossovers. We 
can express the distance between the loci 
of a and /' as being five crossover units long. 
a crossover unit representing that distance 
between linked nonalleles which results in 
one crossover per hundred postmciotic prod- 
ucts (spores, in the present case). Gener- 
ally, when the genes are sufficiently close 
together (as in the present example), cross- 
over frequency (crossover distance) is just 
one half the chiasma frequency, supporting 
our expectation of one chiasma for each 
preceding crossing over. 

Crossover frequency can be measured in 
several ways in Neurospora: 

1. Spores are tested from each sac (two 
to five per sac are sufficient) to determine 
whether or not the sac carries a crossover 
in the region under investigation. In the 
a-b example above, 10% of the sacs would 
have crossovers, 90% would not. Since 
each sac in the 10% group contains four 
spores that are crossovers and four that are 
not, crossover frequency would be 5%. 



TETRAD 
WITH ONE 
CHIASMA 






S 



MEIOTIC 
PRODUCTS 



A 


- B 


A 


• b .— n 


a 


• B 


a 


. '■' — 



Noncrossover 
Crossover 
Crossover 
Noncrossover 



figurl 9—11. Correlation between genetical 
and cytological crossovers. 



2. All the spores from many sacs are 
mixed, then a random sample of spores is 
taken and tested. This method would also 
give 595 recombination with a-b and is sim- 
ilar to the sampling procedure involved in 
determining crossover frequency in animal 
sperm. 

3. One randomly chosen spore from 
each sac is tested; the others are discarded. 
Again. 5% crossovers are obtained. This 
procedure resembles the situation in many 
females (including Drosophila and human 
beings) in which one random product of 
meiosis normally enters the egg and the 
others are lost. 

In the discussion above, no direct correla- 
tion was made between a genetically de- 
tected crossover and a cytologically detect- 
able event involving a particular chromo- 
some region. Such a connection cannot be 
made if both members of a pair of homol- 
ogous chromosomes are identical in cyto- 
logical appearance (as is assumed in Figure 
9-6) because a crossover strand, having ex- 
changed a cytologically identical segment 
with its homolog, appears the same as a non- 
crossover strand. A dihybrid for linked 
genes can be constructed, however, in which 
one homolog differs cytologically from its 
partner on both sides of the loci being tested. 
Such a genetic dihybrid is also cytologically 
dihybrid as specified in Figure 9-11. In 
this case it is possible to collect noncross- 
over progeny and show cytologically that 
they invariably retain the original chromo- 
somal arrangement; crossovers on the other 
hand always show cytologically a new chro- 
mosome arrangement explained by a mutual 
exchange of specific chromosome regions 
between the homologs. :{ 



:! Using this method, genetic crossovers were cor- 
related exactly with cytological crossovers by C. 
Stern (1931) using Drosophila and by H. S. 
Creighton and H. McClintock (1931) using maize. 



Linkage and Crossing Over Between Genes 



127 




Maize workers (/. to r.) at Cornell University in 1929: C. R. Burnham, 
M. M. Rhoades, G. W. Beadle {crouched), R. A. Emerson, and Barbara 
McClintock. 

SUMMARY AND CONCLUSIONS 

The nonallelic genes in a given chromosome are linked and tend to be transmitted 
together to the next generation. Just as intergenic linkage produces an exception to 
independent segregation, crossing over produces an exception to linkage between non- 
alleles and causes linkage to be incomplete. Incomplete linkage proves that a chromo- 
some contains more than one gene. In any given case, the degree to which linkage is 
incomplete — as measured by crossover frequency — is constant and independent both 
of the specific alleles which are present at the two different loci and of the gene com- 
binations that enter the individual forming the gametes. Moreover, reciprocal cross- 
over types are equally frequent. The crossover frequency in different cases is found to 
vary considerably. 

A crossover chromosome is derived from a tetrad in which a crossing over between 
the linked genes showing recombination involves only two of the four strands. For 
closely linked genes, the crossover frequency is one half the frequency with which a 
chiasma or a crossing over occurs between their loci. 

It is hypothesized that crossover frequency is directly related to the distance between 
genes in a chromosome. One unit of crossover distance between genes is defined as 
one crossover per one hundred postmeiotic cells (spores or gametes). Since different 
genes linked to the same gene show different percentages of crossing over with this 
gene, they are presumably different distances from it. 



128 CHAPTER 9 




Curt Stern, in the early 1930's. 



REFERENCES 

Committee on Standardized Genetic Nomenclature for Mice, "A Revision of the 
Standardized Genetic Nomenclature for Mice," J. Heredity, 54:159-162, 1963. 

Creighton, H. S., and McClintock. B., "A Correlation of Cytological and Genetical 
Crossing-over in Zea Mays," Proc. Nat. Acad. Sci., U.S., 17:492-497, 1931. 
Reprinted in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood Cliffs, 
N.J.: Prentice-Hall, 1959, pp. 155-160; also in Great Experiments in Biology, 
Gabriel, M. L., and Fogel, S. (Eds.), Englewood Cliffs, N.J.: Prentice-Hall, 1955, 
pp. 267-272. 

Morgan, T. H.. "Random Segregation Versus Coupling in Mendelian Inheritance," 
Science, 34:384, 1911. Reprinted in Great Experiments in Biology, Gabriel, 
M. L., and Fogel, S. (Eds.), Englewood Cliffs, N.J.: Prentice-Hall, 1955, pp. 
257-259. 

Stern, C, "Zytologisch-genetische Untersuchungen als Beweise fur die Morgansche 
Theorie des Faktorenaustauschs," Biol. Zbl., 51:547-587, 1931. 



QUESTIONS FOR DISCUSSION 

9.1. Distinguish between sex-linkage and the linkage of nonalleles. 

9.2. Does the linkage of two genes prove they are located on the same chromosome? 
Explain. 

9.3. Discuss the advantages and disadvantages of linkage and crossing over with 
respect to the fitness of individuals carrying certain genotypes. 

9.4. In Drosophila, y and spl are X-linked. A female genotypically + +/y spl pro- 
duces sons. If 3% carry either y + or + spl, what are the genotypes and rela- 
tive frequencies of gametes produced by the mother? Is the father's genotype 
important? Explain. 



Linkage and Crossing Over Between Genes 129 

9.5. Name all the processes so far discussed which lead to genetic recombination. 

9.6. Do you think that one of the main principles in this chapter is that chromosomes 
contain more than one gene? Explain. 

9.7. In light of your present knowledge how would you proceed to state a "law of 
independent segregation"? 

9.8. What evidence do you have that crossing over does not involve the unilateral 
movement of one gene from its position in one chromosome to a position in the 
homologous chromosme? 

9.9. Does crossing over always result in genetic recombination? Explain. 

9.10. In what respect do you think the development of the principles of genetics in 
this text would have been affected had the first two pairs of genes (Rr and Yy), 
simultaneously studied in crosses, been linked? 

9.11. Assume that the gene for woolly hair (Chapter 3, p. 38) is located autosomally. 
A nonwoolly-haired, nonhemophilic man marries a woolly-haired, nonhemophilic 
woman. They have a woolly-haired, hemophilic son. Give the genotypes of all 
three individuals, and the genotypes and frequencies of the gametes usually pro- 
duced by the son. 

9.12. How would you defend the conclusion that the point of crossing over is located 
at exactly equivalent positions on the two homologs? 

9.13. What are the relative frequencies of the phenofypes and genotypes expected from 
a mating between two Drosophila: dihybrid vg+/+b? 

9.14. Two dihybrids, a +/+ b, for autosomally linked mutants a and b in Drosophila, 
are crossed. If 2p equals the frequency of noncrossover eggs and 1— 2p equals the 
frequency of crossover eggs, and if p < 0.5, give the relative frequencies of the 
phenotypes expected among the F, of this cross. 

9.15. What result would you expect from 9.14 if the cross were between dihybrids 

a b/+ +? 

9.16. A wild-type Drosophila female, whose father had crossveinless wings and mother 
had yellow body color, is mated to a yellow male. Give the relative frequencies 
of genotypes and phenotypes expected in the F v 

9.17. A mating in Drosophila produces the results shown below. Give the genotypes 
of the parents, and determine which genes are linked and which are not. 

Sons Daughters 

75 wild-type 92 wild-type 

70 yellow body color, white-eyes 75 white 

21 yellow, white, vestigial wings 28 vestigial 

27 vestigial 20 white, vestigial 

2 yellow 

1 white, vestigial 

9.18. What is the relationship in Neurospora between crossing over and first and second 
meiotic division segregation? 

9.19. In tomatoes, the gene for tall ( + ) is dominant to short (s), and the gene for 
smooth epidermis ( + ) is dominant to rough (/-). A cross between two plants 
produces 208 tall smooth, 9 tall rough, 6 short smooth, 195 short rough. Give 
the genotypes of the parents. 

9.20. What is the percentage of crossovers for two loci, in a species in which both 
sexes undergo crossing over with equal frequency, if a mating between identical 
dihybrids (Ab/aB) gives four equally viable classes of offspring, the smallest 
class comprising 1% of all offspring? 



130 < HAPTER 9 

9.21. How would you prove genetically that the last division in a spore sac of Ncuro- 
spora is a mitotic one.' 

9.22. In the absence ot crossing over, could you determine whether the alternatives 
for two different traits were due to a single pair of genes or to two pairs of 
linked genes? Explain. 

9.23. Draw an attached-X chromosome of Drosophila heterozygous both tor y and 
for in. Show the kinds of gametes which could be obtained after: 

(a) No chiasma 

(b) One chiasma between the nonallelic genes 

(c) One chiasma not between the genes mentioned 

9.24. Suppose one member of a long pair of chromosomes in a plant has a large 
knob at one of its ends, and the other has a small knob at the opposite end. 
Suppose, moreover, that there is also a shorter pair of homologs. one member 
terminating with a large knob, and the other at the opposite end with a small 
knob. What combinations and configurations would you expect to readily find 
in the gametes of this individual? 

9.25. What reasons can you present for believing that germ-line crossing over is based 
neither upon premeiotic nor upon postmeiotic events? 

9.26. Calculate the number of crossover units between black body (b) and dumpy 
wings (dp) in the following Drosophila crosses: 

(a) P! pure black X pure dumpy 
P_, Fj 9 9 X black dumpy $ $ 
F-, wild-type 272 

" black' 774 

dumpy 801 

black dumpy 239 

(b) Pj black dumpy X pure wild-type 
P 2 Fj 9 9 X black dumpy $ $ 
F., wild-type 360 

black 103 

dumpy 97 

black dumpy 314 

9.27. What phenotypic results would you expect in 9.26(a) and (b) if the reciprocal 
mating had occurred in P_,? 

9.28. Test statistically the F.> results of 9.26(a) with those expected from independent 
segregation. 

9.29. Test statistically whether the F L , results in 9.26(a) and (b) differ significantly. 

9.30. A trihybrid Aa Bb Cc is test crossed to aa bb cc. The Fj show that the trihybrid 
produced the following gametes: 

29 ABC 21 abc 

235 ABc 215 abC 

210 Abc 239 aBC 

27 AbC 23 aBc 

(a) Which loci are linked and which are segregating independently? 

(b) Write the genotypes of both parents in view of your answer to (a). 

(c) Give the percentage of crossovers wherever applicable. 



Chapter 10 

GENE ARRANGEMENT; 
CROSSOVER MAPS 



I 



n the preceding chapter, the fre- 
quency of crossing over was 
presumed to be dependent upon 
the distance between genes, the interval 
being measured in crossover units. Differ- 
ent genes linked to a given gene were found 
to give different, essentially constant, cross- 
over frequencies or crossover distances. Let 
us now investigate how these different genes 
are arranged spatially. 

Crossover distances can be used to study 
whether linked genes are arranged in some 
regular three-dimensional configuration such 
as a sphere, cube, prism, or some two- 
dimensional one such as a line, circle, or 
triangle. To map the genes on the basis of 
crossover data, that is. make a crossover {or 
linkage) map, it is necessary to determine 
all the crossover distances for a minimum 
of three linked loci, since two points (such 
as those defined by the crossover distance 
between two genes) are not enough to de- 
termine a specific geometrical arrangement. 

Gene Arrangement 

The arrangement of linked loci can be in- 
vestigated with Drosophila. Using the three 
X-linked genes, y (yellow body color), \v 
(white eyes), and spl (split bristles), di- 
hybrid females of the following types 
are obtained: yw/-\ — (-; y spl -\ — |-; and 
wspl/-\ — |-, and each type is test crossed 
with the appropriate double recessive male. 
The corresponding crossover distances are: 
131 



y to w, 1.5; >• to spl, 3.0; and u> to spl, 1.5. 
Since the crossover distance between y and 
spl equals the sum of the crossover distances 
from y to w and from w to spl, a linear ar- 
rangement for these three genes is described, 
namely, y w spl or spl w y. In other words, 
the genetic map based on crossovers is linear. 
If the reasonable assumption is made that 
crossing over is a function of physical dis- 
tance between genes, the genes are also 
linearly arranged in the chromosomes. 

When the positions of a fourth X-linked 
gene and all other X-linked genes are 
mapped relative to the three studied above, 
all are found to be arranged in a linear order 
(Figure 10-1, and page s-16). In such a 
crossover map, y is arbitrarily assigned the 
position, or locus, zero. 

On a standard crossover map for the Dro- 
sophila X, the genes y, w, spl, cv. ct, m, 
and / line up respectively at positions 0, 1.5, 
3.0, 13.7, 20, 36.1, 56.7, and one can see 
that ct and spl are 17 map units apart 
(20 — 3). Since one crossover map unit 
equals one crossover per hundred gametes, 
the dihybrid for spl and ct (Figure 10-2) 
should produce 17% crossovers (8.5% 
+ + and 8.5% splct). However, such a 
result is obtained only under special condi- 
tions. 

The crossover frequency actually observed 
will depend upon several factors. One of 
these is the number of individuals making 
up the sample. In small samples it is very 
likely that, by chance, the observed values 
will deviate considerably in both directions 
from the standard map distance. As the 
size of the sample increases, the observed 
value will more closely approach the stand- 
ard one. Standard distances, therefore, are 
determined only after large numbers of prog- 
eny have been scored. 

The relative viability (see p. 69) of dif- 
ferent phenotypic classes is another factor 
influencing observed crossover frequency. 



132 



CHAPTER 10 



Xorl 




centromere 



0.0 0.1 

Hw svr 



S^. \IHnl 



llw 



.SIT 

pn 
w 
spl 
ec 

hi 

rh 
cv 
rux 
cm 

ct 
sn 



0.8 3.0 6.9 7.5 15.0 18.7 

pn spl bi rb rux cm 

10 unit i of Map Distance 



21 27.7 32.8 38.7 40.7 59.5 

sn Iz ras fw wy fu 

I II I I I 1 1 I I I 



figurh 10—1. Crossover map of the X chromosome of D. melanogaster. 



Name 

yellow body color 

Hairy-wing — extra bristles on wing veins, 
head, and thorax 

scute — absence of certain bristles, espe- 
cially scutellars 

silver body color 

prune eye color 

white compound eyes and ocelli 

split bristles 

echinus- — large and rough textured eyes 

bifid — proximal fusion of longitudinal 
wing veins 

ruby eye color 

crossveinless — crossveins of wings absent 

roughex — eyes small and rough 

carmine eye color 

cut — scalloped wing edges 

singed — bristles and hairs curled and 
twisted 



Kii ro Symbols 
Symboi 



OC 

I 

Iz 

ra.s 

v 

m 

fw 

wy 

s 

8 
sd 

f 
B 
fu 

car 
bb 



Name 

ocelliless — ocelli absent; female sterile 

tan body color 

lozenge — eyes narrow and glossy 

raspberry eye color 

vermilion eye color 

miniature wings 

furrowed eyes 

wavy wings 

sable body color 

garnet eye color 

scalloped wing margins 

forked — bristles curled and twisted 

Bar — narrow eyes 

fused longitudinal wing veins; female 

sterile 
carnation eye color 
bobbed — short bristles 



The phenotypic expression of a + allele is 
usually more viable than that of its mutant 
forms. For example, in Figure 10-2 the 
phenotypically split, cut sons are not as 
viable as the normal (wild-type) sons; al- 
though both types are equally frequent as 
zygotes, the former fail to complete their 
development more often than the latter and, 
therefore, are relatively less frequent when 
the adults are scored. Zygotes destined to 
become either split or cut males are also 
less viable than zygotes destined to produce 
wild-type males. Whenever phenotypes are 
scored after some long developmental pe- 
riod, much of the error due to differential 
viability may be avoided by providing op- 
timal culture conditions. Another way to 
avoid most of this kind of error is to delay 



the scoring of crossovers for one generation. 
The cross is arranged in such a way that 
individuals with the chromosomes to be 
scored have a homologous chromosome con- 
taining the normal alleles of all genes under 
crossover test. Since the progeny of such 
a cross are phenotypically normal, their 
viability will be approximately the same, 
and they can be scored for chromosome 
type from the offspring each produces when 
individually test crossed. For example, the 
female in Figure 10-2 is crossed with wild- 
type males, and the Fi daughters (all pheno- 
typically normal) are individually mated to 
any male. Daughters carrying an X of one 
of the following types: spl +, + ct, -\ — \-, 
spl ct — in addition to a H — \- homolog ob- 
tained from the father — will produce sons of 



Gene Arrangement; Crossover Maps 



133 



the following type respectively: some split, 
but none cut; some cut, but none split; all 
normal; some both split and cut. In this 
way the generation being tested for cross- 
over frequency is largely protected from 
differential viability, its genotypes being 
scored in the next generation. For map- 
ping and other purposes, the extra labor 
entailed by this method is often justified. 

Variability in crossover frequency may be 
due also to factors — such as temperature, 
nutrition, age of the female, and presence 
of specific genes — which influence the very 
process of crossing over. 

To better understand the relationships be- 
tween crossover maps and chiasmata, con- 
sider the properties of an over-simplified 
model (Figure 10-3). Assume that a chro- 
mosome (ignoring the centromere) is com- 
posed of five equally long regions, the ends 
of each marked by a gene; that each tetrad 
of this type contains one, and only one, 
chiasma following a crossing over; and that 
this chiasma can occur in a random posi- 
tion among these segments. For the hexa- 
hybrid shown in Figure 10-3, the chance 
the chiasma will occur in the a-b region is 
20%; out of each 25 tetrads (producing 
100 haploid meiotic products), five or 20% 
will have the chiasma in the a-b region. 
These five will produce 10 crossover and 10 
noncrossover strands. Adding the latter 10 
and the 80 noncrossover strands from the 
remaining 20 tetrads, gives 90 noncrossover 
strands. For this region, therefore, 20% of 
tetrads have a chiasma and 10% of haploid 
meiotic products are crossovers as explained 
in Chapter 9. Similarly, in the b-c region 
10% crossovers would be noted. The 
chiasma would occur in the a-c region 40% 
of the time, and 20% of all haploid meiotic 



figure 10-3. Crossover consequences of a 
single chiasma. 



p i *+l? 9 x ANY O* 

F Sons: spl + / Y 41.5 % 

+ ct / Y 41.5 % 

+ + / Y 8.5 % 

spl ct / Y 8.5 % 

figure 10-2. Crossover frequency for two X- 
linked genes in Drosophila. 





MEIOTIC 40 
PRODUCTS 

40 





45 



45 



PERCENT OF ALL PRODUCTS 




134 



CHAPTER 10 



I 




FIG1 R] 10-4. Chromatid recombinations pos- 
sible in a double chiasmata. {See text for 
details. ) 



products would be crossovers relative to the 
markers a and c. Since the sum of the dis- 
tances from a-b and b-c equals the distance 
between a and c measured directly, the genes 
of the model would be aligned linearly, just 
as was observed in the experiment described 
earlier in this chapter. 

In the model proposed above, the pres- 
ence of a chiasma in one region automat- 
ically excludes it from being in some other 
region. Consequently, the chance that a 
chiasma will be found in the a-c region is 
equal to the sum of the separate chances that 
a chiasma will be found in the a-b and b-c 
regions. It is a general rule that the overall 
probability jor the occurrence of any one 
of a series of mutually exclusive events is 
equal to the sum of their separate probabili- 
ties of occurrence. Therefore, the chance 
of a chiasma occurring between a and / is 
(20 + 20 + 20 + 20 + 20)%, or 100%. 
As a result, 100% of the tetrads have one 
crossing over (that is, one chiasma) which 
produces 50% crossovers, and the model 
chromosome has 50 map units. 

The oversimplification of this model can 
be appreciated by remembering that a given 
tetrad usually contains more than one chi- 
asma. This prompts us to ask: When a 
tetrad contains two or more chiasmata, which 
strands are involved in the exchanges? 

To answer this question, let us specify the 
strands in a tetrad as 1,2, 3, 4, where 1 and 
2 are the sister strands carrying the normal 
alleles and 3 and 4 the sister strands carry- 
ing the recessive alleles (Figure 10-4). If 
one chiasma involves an exchange between 



nonsister strands 2 and 3 in the a-b region, 
a second chiasma, involving nonsister strands 
in the b-c region, can result from any one 
of four exchanges: 2 with 3; 2 with 4; 1 
with 3; 1 with 4. The positions of these 
chiasmata are indicated in Figure 10-4. 
The four types of single chiasma in the b-c 
region together with the single chiasma in 
the a-b region form double chiasmata of 
three types: 

2-strand (the same two strands exchange 
in both chiasmata); 

3 -strand (one of the two strands of the 
first chiasma exchanges in the second, there 
being two ways this double chiasmata can 
occur); 

and 4-strand (those strands which do not 
exchange in the first chiasma, exchange in 
the second). 

Let us examine the genetic consequences 
of these four nonsister types of double chi- 
asmata (shown separately at the left of Fig- 
ure 10-5). The middle column shows the 
meiotic products of each, and the right 
column indicates whether these products are 
noncrossovers, single crossovers, or double 
crossovers for the a-b-c region. From 2- 
strand double chiasmata, two of the four 
meiotic products are genetic noncrossovers 
(+ + + and a be), and two are double 
crossovers (+ b + and a + c). The dou- 
ble crossovers, or "doubles" as they are 
called, are characterized by a change in the 
position of the middle gene relative to the 
end genes. A 3-strand double chiasmata 
produces one double crossover, two single 
crossovers (in each, the position of one end 



Gene Arrangement; Crossover Maps 



135 



gene is changed relative to the other two 
genes), and one noncrossover. The 4- 
strand double chiasmata yields four single 
crossover strands. Note that each type of 
double chiasmata produces some strands 
with a new genetic combination, that is, 
crossover strands; each of the three differ- 
ent types also produces a characteristic pat- 



tern of noncrossover and crossover types. 
Moreover, the genetic products obtained 
from each type of double chiasmata differ 
from those obtained from a single chiasma 
(which produces two noncrossovers and two 
"singles"). 

In view of the preceding discussion, it 
should be possible to learn, from the geno- 



CHIASMATA 



MEIOTIC PRODUCTS CROSSOVERS 




2-STRAND 



2 Doubles 




£ 
J 



a / x _ b _\/_ _ c_ _ 



»\ 



2 b _/ \ __»■_. 

3-STRAND 




c 1 Double 
c 2 Singles 
+ 1 Noncrossover 



Ia_ _b_ 



+ 


b 

+ 


+ 


1 Double 


a 


+ 


2 Singles 


a 


b 


c 


1 Noncrossover 



3-STRAND 



FIGURE 10-5. 

Double nonsister 
chiasmata types 
and their genetic 
consequences. 






+ + 

y 



b !\ 




4 Singles 



4-STRAND 



i:m 



( II \ I'M K Id 



types of the meiotic products, the relative 

frequency with which the tour types oi 
double chiasmata occur. It all four types 
occur with equal frequency, the strands 
forming one chiasma would he unaffected 
by those which form an adjacent chiasma. 
Indeed, experiments with Neurospora reveal 
that all four types c\o occur — in some, the 
four types occur with equal frequency. For 
our purposes, we can accept the view that 
there is usuallj no chromatid interference 
in chiasma formation; in other words, the 
particular nonsister chromatids forming a 
chiasma arc not influenced by nonsister 
strands which may or may not form a chi- 
asma in an adjacent region. Thus, nonsister 
strands crossing over in two different re- 
gions of the same tetrad are independent. 

Does the occurrence of one chiasma in- 
fluence the probability that a second chiasma 
will occur in the same tetrad, even though 
when both chiasmata occur there is no chro- 
matid interference? 

Suppose that in the genetic system of Fig- 
ure 10-4, each of the two regions under 
observation has a 20% chance of forming 
a single chiasma. If the occurrence of a 
chiasma in the a-b region is independent of 
a chiasma in the b-c region, then, of all 
tetrads, 20% of the 20% with an a-b chi- 
asma will simultaneously have a b-c chiasma; 
that is, 4% will contain double chiasmata. 
(According to the previous discussion, this 
4% will be composed of the four nonsister 
types in equal frequency.) It is a general 
rule that the overall probability for the si- 
multaneous or consecutive occurrence of two 
or more events of independent origin is equal 
to the product of their separate probabilities 
of occurrence. 

If 4% double chiasmata were actually ob- 
served in the a-c region, one would conclude 
there was no chiasma interference (or better 
still, no crossing-over interference); that is, 
the formation of one chiasma would not 



affect the formation o\' another in an adja- 
cent region. It. on the other hand, only 
29? double chiasmata were observed, this 
would mean that some chiasma interference 
had occurred. 

The degree o\' chiasma interference can 
be written as 



double chiasmata observed 0.02 
double chiasmata expected 0.04 



= 0.50 



This fraction, called the coefficient of coin- 
cidence, expresses the frequency with which 
the coincidence of two chiasmata is actually 
obtained. Consequently, a coefficient of co- 
incidence equal to zero would mean that one 
chiasma completely prevented the other from 
occurring; whereas a value of one would 
mean that the one chiasma had no effect 
at all on the occurrence of the other. 

In practice, however, because of the errors 
involved — particularly those stemming from 
chiasma terminalization (see p. 22) — one 
does not usually determine the frequencies 
and positions of double chiasmata cytolog- 
ically. Can we use the frequency of ge- 
netically-detected double crossovers as an 
alternative for measuring chiasma or cross- 
ing-over interference? 

We can be sure that each double cross- 
over observed has resulted from multiple 
crossing over or chiasmata. The expected 
frequency of double crossovers in the a-b-c 
region of the example can be calculated in 
the following way: since each region {a-b 
and b-c) has a 0.2 chance for one crossing 
over, the chance for a double crossing over 
is 0.2 times 0.2, or 0.04. (If, as before, the 
coefficient of coincidence were 0.5, one 
would expect 0.02 tetrads to have double 
crossing over.) Recall that the double 
crossing over can occur in four ways and 
can involve 2, 3, or 4 strands of a tetrad. 
If these alternatives occur with equal fre- 
quency, only one quarter ( % 6 ) of all mei- 
otic products from double crossing over will 



Gene Arrangement; Crossover Maps 



137 



appear as double crossovers (Figure 10-5). 
Since the remaining three quarters ( 1 tig) 
of the meiotic products are noncrossovers 
or single crossovers, they are not useful in 
identifying the occurrence of double cross- 
ing-over events, because they could have 
been produced in tetrads of other types, for 
example, those having single or no crossing 
over. Accordingly, a frequency of 0.04 
double crossing over would lead us to ex- 
pect a frequency of .01 double crossovers; 
and a frequency of only 0.005 would actually 
be detected were the coefficient of coinci- 
dence 0.5. In this way, the coefficient of 
coincidence can be determined from double 
crossovers observed divided by the double 
crossovers expected. 

There is another, perhaps simpler, way to 
calculate the expected frequency of double 
crossovers. In our example above, the 
chance a crossing over will occur is 0.2, and 
the chance that a given strand will be a 
crossover, 0.5. The chance that both will 
occur once is 0.1, and that both will occur 
twice is 0.1 times 0.1 or 0.01. That is, the 
expected chance that a given strand will be 
a double crossover is one percent. Accord- 
ingly, the frequency of observed single cross- 
overs in the a-b region multiplied by the 
frequency of observed single crossovers in 
adjacent b-c region equals the expected fre- 
quency of double crossovers (one in each 
region). In practice, therefore, one may 
readily determine the coefficient of coinci- 
dence from double crossovers. 

Generally the coefficient of coincidence is 
negligible — equal to zero for all practical 
purposes — for short map distances and be- 
comes larger with increased distance. This 
relation suggests that a tetrad in which one 
crossing over occurs is somehow precluded 
from having a second one occur close by, 
with this restriction diminishing as the dis- 
tance to the second region increases. In 
Drosophila, for example, the coefficient of 



coincidence is zero for distances up to 10-15 
map units and, consequently, no double chi- 
asmata (or no double crossovers) occur 
within such distances. As the distance in- 
creases beyond \5 map units, however, the 
coefficient gradually increases to 1, at which 
point nothing interferes with the formation 
of double chiasmata. In two equal-armed 
chromosomes there does not seem to be 
chiasma interference across the centromere. 

If each tetrad has only a single chiasma, 
the maximum frequency with which the end 
genes recombine relative to each other is 0.5. 
What happens to the frequency of recom- 
bination for the end genes when the chro- 
mosome has double chiasmata? 

If each tetrad has two chiasmata, one 
might think that the end genes would form 
new combinations with a frequency greater 
than 0.5. Examination of Figure 10-5 re- 
veals, however (each type of double chias- 
mata being equally probable), that on the 
average eight products (single crossovers) 
will carry a new combination with respect to 
one end gene, and eight products will not. 
Of the latter, four will be noncrossovers and 
four, double crossovers in which the middle 
gene has changed position relative to the 
end genes. Therefore, even if every tetrad 
has double chiasmata, the maximum fre- 
quency of recombination for the end genes 
is 0.5. 

When four loci are studied and three chi- 
asmata occur in each tetrad — one in each 
region — one finds that for every 64 meiotic 
products, 32 are recombinational for the 
end genes and 32 are not. For cases where 
four or more chiasmata lie between end 
genes, the frequency of meiotic products 
bearing odd numbers of crossover regions 
is easily calculated to be 0.5. In each of 
these cases the gene at one end is shifted 
relative to that at the other. However, the 
remaining strands contain either even num- 
bers of crossover regions (which do not 



138 



CHAPTER 10 



+ + + 0.31 

a b c 0.31 

+ b c 0.14 

a + + 0.14 

+ + c 0.01 

a b + 0.01 

+ b + 0.04 

a + c 0.04 



+ + + 0.31 

a c b 0.31 

+ c b 0.14 

a + + 0.14 

+ c + 0.01 

a + b 0.01 

+ + b 0.04 

a c + 0.04 



1.00 



1.00 



figure 10-6. Determination of gene order 
from a test crossed trihybrid. 



cause the genes at the two ends to shift 
relative to each other) or are noncrossovers. 
Accordingly, the maximum frequency of re- 
combination of 0.5 holds for the endmost 
genes (and, therefore, of course, for any 
genes between them). 

If two genes in a chromosome are suffi- 
ciently far apart, the frequency with which 
they undergo recombination will be near 0.5. 
Since a recombination frequency of 0.5 
means that nonalleles are independent in 
their segregation, one cannot conclude from 
such a recombination frequency that non- 
alleles are on the same chromosome. Ac- 
cordingly, two pairs of genes that show re- 
combination frequencies near 0.5 can be 
either far apart in the same pair of homologs 
or located in different pairs of homologs. 
However, if two nonalleles segregate inde- 
pendently but are both linked to a third 
nonallele, all three are linked to one an- 
other. 



Whenever the number of gene pairs in- 
vestigated is considerably larger than the 
Dumber of chromosome pairs, the number 
of groups of linked genes equals or ap- 
proaches the number of chromosome pairs. 
The result is a limitation in the number 
of linkage groups, the maximum number 
equalling the haploid chromosome number. 
(Examination of the linkage groups of the 
garden pea now reveals that two of the first 
seven gene pairs studied ' arc in the same 
linkage group although a considerable dis- 
tance apart. The initial recombination data 
were sufficiently meager for acceptance of 
the hypothesis that the genes were segregat- 
ing independently.) 

The sequence of three linked genes can 
be determined from the results of a single 
cross. Suppose the trihybrid -+- -+- + /a b c 
is test crossed, and the frequencies of the 
various phenotypes in the progeny are those 
shown at the left in Figure 10-6. These 
values, we remember, represent the fre- 
quencies of the corresponding genotypes in 
the gametes of the trihybrid. The middle 
gene in the actual sequence is the one which 
switches least often from the original gene 
combinations (+ -\ — h and a be), because 
only the middle one requires two chiasmata 
for its switch. Consequently, this gene is 
identified as c, and the actual gene order is 
acb (or be a). This reasoning may be 
easier to follow if the data are examined 
with the genes listed in their correct order, 
as shown at the right in Figure 10-6. 

The frequency of observed crossovers be- 
tween the a and c loci is 0.30; between c 
and b it is 0.10. Between a and b the fre- 
quency of single crossovers is 0.36. Cross- 
over frequency between a and b, however, 
also includes double crossovers. Since each 
double crossover represents two single cross- 
overs between the end genes, the frequency 

1 By G. Mendel. 



Gene Arrangement; Crossover Maps 



139 



of double crossovers, .02, must be doubled 
and added to the frequency of single cross- 
overs to obtain the total crossover frequency 
between a and b. The genetic map based 
on crossover frequency becomes linear 
(0.30 + 0.10 = 0.36 + 0.04), therefore, 
when double crossovers are taken into ac- 
count. The expected frequency of double 
crossovers is 0.3 times 0.1 or 0.03, so that 
the coefficient of coincidence in this case is 
0.02 0.03, or 0.66. ( In the y w spl example 
discussed earlier in this chapter, the longest 
region, y-spl, was too short for double cross- 
ing over.) 

Would it be satisfactory to use the data 
in Figure 10-6 to construct a standard link- 
age map for the distances between these 
genes, assuming that large numbers of prog- 
eny had been scored and standard experi- 
mental conditions had been used? For this 
purpose, the observed distance from c to b 
is acceptable since only a single chiasma can 
occur in such a short interval. The situa- 
tion is otherwise for the a-c region, however, 
which is 30 map units long in the present 
experiment. Double chiasmata are expected 
to occur under these circumstances, yet the 
absence of genetic markers between a and c 
prevents their identification. Therefore, the 
standard map distance for a-c must be longer 
than 30 map units (and a-b longer than 40) . 
Note that the identical error foreshortens the 
a-c and the a-b distances; therefore, for the 
distances observed, {a-c) plus (c-b) is 
equal to (a-b). Whether or not the chro- 
mosome is genetically marked so that all 
multiple crossover strands are detected, the 
correct order of three linked genes can al- 
ways be determined, provided that two are 
not 50 map units away from the third. 

It should now be clear why the crossover 
frequencies observed for large distances are 
less than the standard map distances, and 
why the standard map distances are always 
obtained by the summation of the short dis- 



tances in which only a single chiasma can 
occur. 

Although end genes can show at most 
50% recombination, the length of the cross- 
over map may exceed 50 units. For exam- 
ple, if a given pair of homologs contains an 
average of two chiasmata in each tetrad (see 
Figure 1 0-5 ) , a total of 1 00 crossovers will 
occur among 100 meiotic products, and the 
map length will be 100 units even though 
the end genes will have recombined 50% of 
the time. In fact, it can be predicted that 
the length of the standard map is equal to 
fifty times the mean number of crossing-over 
events (or chiasmata) per tetrad. 

Crossover Maps 

Utilizing crossover frequency, genetic maps 
cf chromosomes have been made for a num- 
ber of multicellular organisms. Figures 10-7 
through 10-10 give the linkage maps for a 
considerable number of genes in man, mouse, 
maize, and Neurospora. 



29 



G6PD Deutan Hemophilia A 
► e5->|«— 12 • 



I 



3 



38- 
-41 



V 



figure 10-7. Tentative linkage map of a 
segment of the human X chromosome. The 
numbers given are the values for the map 
distance found in five separate studies. The 
loci mapped are the Xg {blood group) locus, 
the G6PD (glucose-6-phosphate dehydrogen- 
ase) deficiency locus, the deutan (green) color- 
blindness locus, the classic hemophilia locus. 
(Courtesy of V. A. McKusick. From Human 
Genetics, 1964, Prentice-Hall, Inc., Englewood 
Cliffs, N.J.) 



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I I n II I I I I ■ I ■ III 



Gene Arrangement; Crossover Maps 141 





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Gene Arrangement; Crossover Maps 143 



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144 



CHAPTER 10 



+ trosl 
I 
*mt-2 

ieu-3 
leu-4 
moling 



II 



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jx 



YI 

chol-2 



YH 



-cauliflower 



pyr4 



ad 5 



type 



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suc- 



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liny 



row- IT;^ 

""LIS 



crisp 
visible ■*■ 



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■thil 



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nit-l 



lhre-2 



lboltoon iCumb i-lh.4 pd^ipy^o }"'»* 



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arom-l 



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inrp*i: 



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■leu 



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.. ad -7 



figure 10-10. The 
linkage groups of 

Ncurospora crassa. 
/ he relative \i:e\ and 
centromere positions 
( open circles) of the 
chromosomes are shown 
at the lower right. 
I he chromosomes and 
linkage groups are 
numbered independ- 
ently. Map distances 
are only approximate. 
( Courtesy of Black well 
Scientific Publications, 
Oxford, from 
Fungal Genetics 
by J. R. S. Fincham 
and P. R. Day, 1963.) 



asp 
plug 



ii i AAA 

I 2 3 4 5 6 7 



Key to Genetic Symbols 
Most descriptions refer to nutritional requirements of the corresponding mutants. 

Symbol Name 

a/ A mating type 

ac acetate 

ad adenine 

al albino (white conidia) 

am Q-amino nitrogen (de- 
ficient in glutamic 
dehydrogenase) 

amyc amycelial (an ex- 
treme morpholog- 
ical variant) 

arg arginine 

arom aromatic (tyrosine -f 
phenylalanine + 
tryptophan + p- 
aminobenzoic acid ) 

asp asparagine 

aur aurescent (probably 
an allele of al-l ) 

can canavanine resistance 

chol choline 

col colonial morphology 

cys cysteine 



Symbol 


Name 


Symbol 


Name 


cyt 


altered cytochrome 


prol 


proline 




system 


pyr 


pyrimidine 


for 


formate 


rib 


riboflavin 


hist 


histidine 


ser 


serine 


hs 


homoserine 


sfo 


sulphonamide- 


inos 


inositol 




requiring 


iv 


isoleucine + valine 


su 


suppressor 


leu 


leucine 


sue 


succinate or other 


lys 


lysine 




Krebs cycle inter- 


me 


methionine 




mediate 


nic 


nicotinic acid 


tin 


thiamin 


nit 


non-nitrate utilizing 


thr 


threonine 


nt 


nicotinic acid or 


tryp 


tryptophan 




tryptophan 


tyr 


tyrosine 


OS 


sensitive to high os- 


T 


modified tyrosinase 




motic pressure 




structure 


ox-D 


deficient in D-amino 


vol 


valine 




acid oxidase 


vis 


visible (used for vari 


pah 


p-aminobenzoic acid 




ous morphological 


pan 


pantothenic acid 




mutations) 


pdx 


pyridoxine 


ylo 


yellow conidia 



Gene Arrangement; Crossover Maps 145 




Alfred H. Sturtevant {in 1945). 

SUMMARY AND CONCLUSIONS 

Using crossover frequency as an indication of distance between loci, linked genes are 
found to be arranged linearly. Observed crossover frequencies fluctuate because of 
variations in sample size and factors (temperature, age, nutrition, genotype) which 
affect either the crossing-over process itself or exert their influence after crossing over 
(differential viability). Standard crossover maps are made under standard (optimal 
for crossing over) conditions. 

One crossing over (or chiasma) can interfere with the occurrence of another in the 
same tetrad. This crossing-over (or chiasma) interference diminishes as the distance 
between the two regions increases. When double crossing over occurs, the chromatids 
that exchange in one crossing over generally have no influence upon which chromatids 
exchange in the other. Consequently, there is usually no chromatid interference. 

Recombination with respect to end genes is 50%, maximally, no matter how many 
chiasmata occur per tetrad. Although the order of linked genes is easily determined 
by test-crossing trihybrids, the distance between two marked loci is underestimated 
when multiple crossovers between them are not detected. 

REFERENCES 

Barratt, R. W„ Newmeyer, D., Perkins, D. D., and Garnjobst, L., "Map Construction 
in Neurospora crassa," Advances in Genetics, 6:1-93, 1954. 

Emerson, R. A., Beadle, G. W., and Fraser, A. C, "A Summary of Linkage Studies 
in Maize," Mem. Cornell Univ. Agr. Sta., No. 180, 1935. 



I H, ( H M-l IR 10 

1 ou is. K. R.. and John. B.. Chromosome Murker. 1 omlnn. I. and A. Churchill. 1 td., 
1963. 

Sturtevant, A H.. "The 1 inear Arrangement ol Six Sex-Linked I actors m Drosophila. 
.^ Shown bj I hen Mode oi Association," I. Exp. Zool., 14:43 59, 1913. Re- 
printed in Classic Papers in Genetics, Peters. I. A. (Ed.). Englcwood Cliffs, N.J.: 
Prentice-Hall. 1959, pp. 67-78. 

See Supplement II. 

QUESTIONS FOR DISCUSSION 

10.1. Does the linear arrangement ol the -cues offer anj evidence for or against the 
hypothesis that chromosomes are carriers of genes? Explain. 

10.2. How many gene pairs must he heterozygous lor you to detect a single and a 
douhle crossover in Drosophila? In Neurospora? 

10.3. Suppose a pair of homologs in Neurospora have the genotypes A B/a b. Draw 
an eight-spore ascus derived from a cell that had: 

(a) No chiasma hetween these homologs 

(h) One chiasma between the centromere and the gene pair closest to it 

(c) One chiasma between the two pairs of genes 

(d) One two-strand double chiasmata between the two pairs of genes 

10.4. What are the advantages of Neurospora over Drosophila as material for genetic 
studies? 

10.5. Under what conditions would segregation ol a pair ol' alleles occur during the 
first meiotic division? The second meiotic division? 

10.6. What indications might you have that differential viability plays a role in modify- 
ing experimentally obtained crossover distances? 

10.7. How many linked loci must be hybrid in a Drosophila individual and a Neuro- 
spora individual to determine from crossover data whether these loci are ar- 
ranged linearly? Explain. 

10.8. A test cross proves that one of the parents produced gametes of the following 
genotypes: 42.4% PZ, 6.9% Pz; 7.0% pZ\ and 43.7% pz. List all the genetic 
conclusions you can derive from these data. 

10.9. The trihybrid Aa Bb Cc is test crossed to the triple recessive, aahbec, and the 
following phenotypic results are obtained: abc 64; abC 2: aBc 11; aBC 18; AbC 
14; Abc 17; ABc 3; ABC 71. 

(a) Which of these loci are linked? Why? 

(b) Rewrite the genotypes of both parents. 

(c) Determine the observed map distances between all the different pairs of 
linked genes. 

10.10. A Drosophila female with yellow body color, vermilion eye color, and cut 
wings is crossed with a wild-type male. In F t all females are wild-type and 
males are yellow, vermilion, cut. When the F, are mated to each other, the F L , 
are phenotypically as follows: 

1781 wild-type 1712 yellow, vermilion, cut 

442 yellow 470 vermilion, cut 

296 vermilion 265 yellow, cut 

53 cut 48 yellow, vermilion 

Construct a crossover map for y, v, and ct from these data, giving the map 
distances between loci. 



Gene Arrangement; Crossover Maps 147 

10.11. Describe the practical significance of the (act that linked genes are arranged 
linearly. 

10.12. How can you determine the position of a centromere in a linkage group of 
Neurospora? 

10.13. Under what conditions are all eight ascospores from a single sac detectable 

crossovers? 

10.14. In corn, virescent seedlings, glossy seedlings, and variable steriles are due to 
three recessives — \\ gl, and va, respectively. G. W. Beadle test crossed a tri- 
hybrid and obtained the following phenotypic results: 

235 normal 270 v gl va 
40 va 48 v gl 

60 v 62 gl va 

7 gl 4 v va 

Construct a crossover map for these three genes, giving the map distances be- 
tween loci. 

10.15. What effect do undetected multiple crossover strands have upon gene sequence 
of marked loci? Observed map distance for marked loci? 

10.16. What evidence can you present that crossing over involves segments of homologs 
rather than individual loci? 

10.17. Explain the following statement: The frequency of first division segregation of a 
gene pair in Neurospora is inversely related to its distance from the centromere. 

10.18. How can you convert the percentage of asci showing second division segregation 
into map distance from the centromere? 

10.19. In Neurospora, the mutant ag cannot synthesize the amino acid arginine, and the 
mutant th fails to synthesize the vitamin thiamin. Since both substances are 
needed for survival, explain how a stock pure for ag th is maintained and tested 
for the presence of these mutants. 

10.20. A dihybrid for the mutants described in Question 10.19 produces asci having the 
following spore orders: 

Spores 



asci 


1 +2 


3 + 4 


5 + 6 


7 + 8 


24 


ag th 


ag th 


++ 


+ + 


27 


+ + 


++ 


ag th 


ai> th 


26 


ag+ 


ag+ 


+ th 


+th 


23 


+ th 


+th 


ag+ 


ag+ 



Discuss the positions of these loci with respect to each other and their centro- 
mere(s). How would you determine the genotypes present in these spores? 

10.21. Map mutant a relative to its centromere when Neurospora heterozygous for this 
mutant produces asci having the following spore orders: 

Spores 



asci 


1 +2 


3 + 4 


5 + 6 


7 + 8 


44 


a 


a 


+ 


+ 


48 


+ 


+ 


a 


a 


2 


a 


+ 


a 


+ 


3 


a 


+ 


1 


a 


2 


+ 


a 


+ 


a 


1 


+ 


a 


a 


4- 



148 



CHAPTER 10 



10.22. Suppose asci from a given Neurospora cross had spores in the following relative 
order: 

Spores 



% 


1 I + 2 


3 + 4 


5 + 6 


7 ! 8 


asci 


or i 7 + 8 


5 + 6 


3 + 4 


1 +2 


92 


xy 


XV 


+ + 


+ + 


2 


xy 


++ 


xy 


+ + 


2 


xy 


++ 


++ 


xy 


1 


xy 


x+ 


+y 


++ 


1 


xy 


x+ 


++ 


+y 


1 


x+ 


xy 


+y 


++ 


1 


X+ 


xy 


++ 


+v 



(a) Are x and y linked? 

(h) If not linked, give the crossover distance from its centromere for each, [f 

linked, construct a crossover map for x and y and their centromere. 
(c) Are any double crossovers involved? Why? 

10.23. The accompanying photograph (courtesy of R. G. Isaacson) shows asci of the 
fungus Sordaria fimicola. The asci are in various stages of maturity, the most 
mature containing dark ascospores. What genetic conclusions can you draw 
knowing that all the asci shown are products of the same parental genotype? 




Chapter 11 

CHANGES INVOLVING 
UNBROKEN CHROMOSOMES 



E 



xcept for Chapters 6 and 8 
the preceding chapters sought 
to determine the characteris- 
tics of the genetic material through the oper- 
ation of genetic recombination. This opera- 
tion revealed the existence of different re- 
combinational units of the genetic material, 
which in order of size, include the genome, 
the chromosome, and the genes in a chro- 
mosome — the smallest recombinational unit 
being the recombinational gene. 

This chapter begins a study of the genetic 
material through the operation of mutation. 
We shall be especially interested in learn- 
ing the extent to which the genetic material 
can be divided into mutational units, always 
remembering that the recombinational and 
mutational units may or may not be mate- 
rially identical. 

We have been able to learn the recom- 
binational properties of the gene only be- 
cause it exerts a detectable phenotypic effect, 
and because it exists in an alternative state. 
One can readily see that if a gene were pres- 
ent in the same form in all organisms, it 
would not be detectable, since all individuals 
would have the same genotype and, there- 
fore, the same range of phenotypic expres- 
sion. In other words, the genes detected 
thus far in this book were only those that 
occur either in different numbers in different 
individuals, or have an alternate allele, or 
both, provided that such a genetic difference 
produced a detectable phenotypic change. 
149 



A great deal of genetic variation of this 
kind exists among living organisms. We 
have seen that some of the phenotypic varia- 
tion attributed to genes is actually due to 
sexuality which by segregation, independent 
segregation, crossing over, and fertilization 
produces new combinations of already-pres- 
ent genes. These mechanisms of recombina- 
tion shuffle the genes, just as shuffling a deck 
of playing cards produces the great variety 
of card combinations. 

Detecting Mutations 

We would like to learn two things concern- 
ing genetic differences; namely, what they 
are, and how they are produced. To do this 
we must first have some way to distinguish 
between a mutant — a really new genetic 
form produced by the process of mutation — 
and a recombinant for already-existing genes. 
We can use an example in Drosophila to 
illustrate how this distinction may be made. 
Suppose (as was the case at one time) none 
of the flies in laboratory strains has an ap- 
pendage on the anterior-dorsal part of the 
thorax. Then a single fly occurs with an 
appendage in this region (Figure 11-1) and, 
when crossed with the wild type of a dif- 
ferent strain (outcrossed), this trait appears 
in approximately one half of the progeny. 
How is the new phenotypic variant (He.x- 
aptera) to be explained? 

If the culture conditions had not changed. 
Hexaptera could not be due to environmental 
factors alone. Could Hexaptera result from 
a new combination of already-existing ge- 
netic units? It could not be due to the inter- 
action between two particular alleles al- 
ready present in the population which hap- 
pened to combine in the same zygote at 
fertilization, for such a combination would 
have to be rare and, following segregation, 
this phenotype would not be expected to 
appear in any appreciable number of the 
progeny of the outcross. Moreover, it could 



150 



( II M'TI R 1 1 



not be due to the rare combination of two 
already-existing unlinked nonalleles since, at 
most, onlj one quarter of the progenj would 
have the novel phenotype. Consequently, 
neither segregation nor independent segrega- 
tion could be associated with ttie appearance 
or nonappearance o\ Hexaptera. The new 
phenotype might have appeared alter a rare 
crossing over between two \er\ close loci 
brought two previously separated nonalleles 
into the same chromosome. Once produced, 
this new combination of linked genes would 
almost always remain intact and be trans- 
mitted to one half of the progeny. How- 
ex er. suppose also that the parents' chromo- 
somes were suitably marked witli genes, and 
it was found that the chromosome region. 
essential for the production of the new phe- 




I IGl Rl 11-1. 
melanogaster. 
1949.) 



The Hexaptera phenotype in D. 
(from Genetics, vol. 34, p. 13, 



notype, was o\' a noncrossovcr type. In 
such a case, crossing over would not explain 
the results either. 

The only reasonable remaining explana- 
tion would be that a novel change, a muta- 
tion, occurred in the genetic material. We 
sec. therefore, that when the mutant pro- 
duces a dominant phenotypic effect it is not 
too difficult to determine whether a novel 
phenotype is due to mutation rather than to 
genetic recombination. 

In the case of a dominant mutant, only 
one parent needs to have a specific genotype 
to produce a dominant mutant trait in the 
progeny; no particular genetic recombina- 
tion is a prerequisite. In other cases, the 
novel phenotype appears in the progeny only 
when both parents have specific genotypes, 
and genetic recombination is required for 
its appearance. Note that the detection of 
a completely recessive autosomal mutant 
gene is postponed for the number of genera- 
tions required for two heterozygotes to mate 
and produce a mutant homozygote. Before 
a recessive mutant becomes homozygous, 
many generations may elapse, during which 
time the mutant allele may become relatively 
widespread in the population in heterozygous 
condition. When the genotype of the pop- 
ulation is uniform or is known, it may be 
possible to trace back to the origin of a new 
recessive mutant. If, however, the popula- 
tion genotype is not known, it is impossible 
to determine when a recessive mutant first 
arose, and it may be considered — correctly 
or not — one of the genes normally present 
in the population. 

Obviously, the detection of mutants, both 
recessive and dominant, would be made rela- 
tively easy by using pure lines. As men- 
tioned in Chapter 1. suddenly-appearing 
phenotypic variants which are due to muta- 
tion and not to environmental fluctuation are 
occasionally found in pure lines of self- 
fertilizing bean plants. When completely 



Changes Involving Unbroken Chromosomes 



pure lines cannot be obtained because self- 
fertilization does not occur, detection of 
mutations is facilitated by knowing the pre- 
existing genotypes. 

Although we have seen how a phenotypc 
is proved to be the result of a mutation, we 
have not determined the basis for the genetic 
change involved. The change could con- 
ceivably encompass as much as an entire 
genome, or as little as the genetic material 
in a single gene locus. The latter type of 
change may not be detectable cytologically. 
Although a cytological study has not been 
made, genetic studies indicate that the 
chromosomal change associated with the 
occurrence of Hexaptera is submicroscopic. 
Let us now look at mutants known to be as- 
sociated with a gross visible change in chro- 
mosome composition, as detected by either 
genetical or cytological methods, or both, 
and leave for later the consideration of muta- 
tions involving submicroscopic changes in 
chromosomes. 

Heteroploidy 

In the evening primrose, Oenothera, a giant 
type called gigas is found to be a mutant. 
Other Oenothera, like most sexually repro- 
ducing species, are diploid, having two sets 
of chromosomes — one genome contributed 
by each of the gametes. In the gigas type, 
cytological examination shows that there are 
three genomes; that is, the individuals are 
triploid. Studies of other groups of diploid 
plants reveal related types which prove to 
have four genomes (tetraploids), others may 
have six sets {hexaploids) or eight (octa- 
ploids). A chromosomal composition made 
up of an abnormal number of normal chro- 
mosome sets is said to be heteroploid. The 
occurrence of extra whole genomes is called 
polyploidy, a term which is applicable for 
multiples of the haploid number when mono- 
ploidy is the normal condition. Note that 
changes in genome number preserve the 



151 









figure 11-2. Ploidy in Datura (N = 12) 
(silhouettes). 



same ratios that chromosomes (and genes) 
have to each other under normal conditions. 
Such changes are said to be euploid (right- 
fold). 

1. Autopolyploidy 

Different forms of the Jimson weed. 
Datura, carry different numbers of whole 
genomes, or ploidies. 1 Some are haploid, 
others diploid, triploid, or tetraploid. The 
flowers that each of these types produces are 
shown in Figure 11-2, line B (the respective 
seed capsules are shown in line A). Note 
that flower size increases with ploidy. The 
seed capsules illustrated are those which 
might have been obtained had the individual 
under test been fertilized by pollen from a 
diploid — the differences in size being due 
partly to the number of seeds that have set 
or developed. 

Triploid and tetraploid embryos are found 

1 The following is based upon work of A. F. 
Blakeslee and J. Bellinc 



152 



CHAPTER 11 




iiwri: 11-3. Normal {left) and triploid (right) D. melanogaster females. The body 
of the 3N female is slightly larger than the 2N female and also has slightly larger cells. 
(Drawn by E. M. Wallace.) 



in a variety of mammals, and even races of 
some animals are polyploid. For example, 
tetraploids of the following are known: the 
water shrimp, Artemia; the sea urchin. 
Echinus; the roundworm, Ascaris; and the 
moth, Solenobia. Polyploid larvae of sala- 
manders and of frogs also have been ob- 
tained, although races are not formed. Poly- 
ploidy is also found in Drosophila. Female 
Drosophila have been found that are triploid 
(3X + 3 sets of A) (Figure 11-3) and 
tetraploid (4X + 4 sets of A). Somatic 
parts of Drosophila individuals have been 
found to be haploid (IX + 1A set). 

One way that ploidy can increase is by 
the addition of genomes of the same kind 
as those present — by allopolyploidy — as was 
the case with Datura. Autopolyploidy can 
arise several different ways: 

1 . Mitotic anaphase may be abnormal, so 
that the doubled number of chromosomes 
becomes incorporated into a single nucleus 
which thereafter divides normally to produce 
daughter polyploid nuclei and eventually — 
by asexual reproduction — polyploid progeny. 

2. Sometimes two of the haploid nuclei 
produced by meiosis fuse to form a diploid 
gamete which, after fertilization with a hap- 
loid gamete, forms a triploid zygote. (Com- 
plementarily, fertilization of a gamete formed 
without a nucleus may initiate development 
of a haploid.) 



3. Haploid individuals may undergo meio- 
sis and, although this usually results in gam- 
etes containing only part of a genome, a 
complete haploid gamete can sometimes be 
produced which, upon fertilization with an- 
other haploid gamete, forms a diploid zygote. 

By interfering with mitosis and meiosis, 
autopolyploidy can be artificially induced by: 
drugs like colchicine or its synthetic substi- 
tute, colcemide (which destroys the spindle, 
thereby preventing the anaphase movement 
of chromosomes); environmental stresses 
like starvation and cold; or energetic radia- 
tions. 

Some females of Solenobia produce hap- 
loid eggs; others produce diploid eggs. Both 
types of eggs start development without 
fertilization; that is, they begin developing 
parthenogenetically. During development, 
however, nuclei of the respective individuals 
fuse in pairs to establish the diploid and 
tetraploid conditions. In this case, normal 
parthenogenesis leads to normal diploidy and 
tetraploidy. In many other organisms, arti- 
ficially induced parthenogenesis may begin 
haploid development. 

In the case of an ordinarily diploid indi- 
vidual, development as a haploid usually pro- 
duces abnormalities. These must sometimes 
be due to the expression of detrimental genes 
which are not expressed in a diploid because 
their normal alleles are present in homol- 



Changes Involving Unbroken Chromosomes 



153 



ogous chromosomes. However, this is not 
always the case. If chromosome doubling — 
naturally or artificially induced — occurs at 
an early stage, a normal diploid (and homo- 
zygous) embryo may be produced; for ex- 
ample, chromosome doubling has produced 
parthenogenetic salamanders and (female) 
rabbits. In these instances, at least, abnor- 
mal development as a haploid must have its 
basis in quite a different factor — probably 
one involving the surface-volume relation- 
ships within the nucleus and between the 
nucleus and the cytosome. These relation- 
ships are changed when cells that are 
adapted to be diploid are haploid. A sim- 
ilar explanation can be offered for the ob- 
servation that development of triploid and 
tetraploid mouse zygotes ceases after a few 
days, even though initially they have a nor- 
mal mitotic rate. 

Ploidy changes also occur during gameto- 
genesis and fertilization. These and certain 
other examples of ploidy change already dis- 
cussed are normal in various organisms. (A 
ploidy change should be considered muta- 
tional only when it is novel.) Autopoly- 
ploidy can occur as a normal process in a 
portion of a multicellular organism; for ex- 
ample, it occurs normally in certain somatic 
tissues in man such as liver cells. Many 
of the examples of autopolyploidy mentioned 
involve an increase in ploidy which is accom- 
plished by endoreplication; that is, the ge- 
nomic contents replicate and remain in one 
nucleus. In these cases, the daughter chro- 
mosome strands separate to produce an in- 
creased number of separate chromosomes, 
each chromosome in the nucleus proceed- 
ing independently to mitotic metaphase. In 
another consequence of endoreplication, all 
daughter chromosome strands remain syn- 
apsed, so the number of separate chromo- 
somes is not increased. Let us consider an 
example of this condition as found in the 
giant salivary gland cells of Drosophila 
larvae. 



2. Poly ne my 

Recall that the metaphase chromosome in 
the usual cell of Drosophila is rod-shaped 
(see Figure 7-5) and contains chromatids 
each of which is coiled tightly in a series of 
spirals like those in a lamp filament, and 
that during interphase the chromatids un- 
wind. The chromatids in the chromosomes 
of the salivary gland cell nucleus are also in 
an unwound state, perhaps even more so 
than in ordinary interphase, and undergo 
three special changes: 

1. Each chromosome present endorepli- 
cates synchronously a number of times in 
succession, so that one chromosome pro- 
duces two, two produce four, four produce 
eight, and so on. Endoreplication can occur 
at least nine times, so each chromosome can 
produce 512 daughters. 

2. All daughter strands, instead of sep- 
arating, remain in contact with the homol- 
ogous loci apposed, giving the appearance 
of a many-threaded — polynemic or polytenic 
— cable. 

3. The original members of a pair of ho- 
mologous chromosomes are paired at homol- 
ogous loci, demonstrating what is called 
somatic synapsis. Accordingly, a double 
cable is formed which can contain as many 
as 1024 chromosomes. 

When seen under the microscope (Figures 
1 1-4 through 6), these double cables have 
a cross-banded appearance due to differ- 
ences in density along the length of the un- 
wound chromosomes. A band is formed by 
the synapsis of the same dense regions in all 
the strands; in this case, an interband region 
is also formed by the synapsis of correspond- 
ing regions of lesser density (Figure 11-5). 
The pattern of bands is so constant and 
characteristic that it is possible to identify 
not only each chromosome but different re- 
gions within a chromosome (Figure 11-6). 
The giant size of salivary chromosomes, very 
long because they are unwound and thick 



I. 14 



( IIAI'I I K II 



V\ 



' "* X \ 



3L 



'. 



v 

2L 






« '*» 



V x 



% 






kttt* 



} 2R '-'% 






♦ 



3R % 



- 



v* 



*3 




figure 1 1 -4. Salivary gland chromosomes of a female larva of D. melano- 
gaster. (Courtesy of B. P. Kaufmann; by permission of The American Genetic 
Association. Journal of Heredity, Frontispiece, vol. 30, No. 5, Max, 1939.) 



I K.i RE 1 1-5. A hand (at lop) and 
interhand (below) region of a 
stretched Drosophila salivary gland 
chromosome. Photographed with 
the electron microscope at a magni- 
fication of approximately J2.200X- 
Present enlargement is about 
13.000X. (By permission of The 
American Genetic Association. Jour- 
nal of Heredity, vol. 43, p. 231. 
1952.) 

figure 1 1-6. The pair of fourth 
chromosomes as seen in salivary 
gland nuclei ( each homolog is highly 
polynemic) and at mitotic metaphase 

(arrow), drawn to the same scale. 
I By permission of The American 
Genetic Association. C. B. Bridges, 
"Salivary Chromosome Maps," Jour- 
nal of Heredity, vol. 26, p. 62, 
1935.) 



'•/ 



SCALE 



5 A*- 



&%£} 






#l%tviin 



■ .-^•'*-n»v.U-Vt6' «iii 




Changes Involving Unbroken Chromosomes 



155 



because of synapsed polynemes, offers a 
unique opportunity to correlate genetical and 
cytological events. ( It should be noted that, 
as a rule, giant polyncmic chromosomes are 
produced in cells which will never divide 
again. ) 

At any given stage of a cell cycle, most 
of the chromosomal material reacts similarly 
to certain staining procedures and, therefore, 
is called euchromatic (truly or correctly 
colored). Other portions of the chromo- 
somes stain either darker or lighter and are 
said to be heterochromatic. Although het- 
erochromatin may be located at various 
places along a chromosome arm, it is nor- 
mally found adjacent to the centromere and, 
to a lesser extent, near the ends. Hetero- 
chromatin also has the characteristic of being 
less specific in synapsis than is euchromatin, 
different heterochromatic regions located in 
the same chromosome, its homolog, or in 
nonhomologous chromosomes often being 
found synapsed. In the giant salivary gland 
nuclei of Drosophila larvae, the heterochro- 
matic regions nearest the centromeres of all 
chromosomes synapse to form one mass, 
called the chromocenter. This is the center 
from which the double cables radiate in 
Figure 1 1-4 and at the left of Figure 1 1-6. 
Also, the heterochromatic regions nearest 
the ends are sometimes found synapsed with 
other heterochromatic regions, especially the 
chromocenter. In squashing the nuclei to 
separate and flatten the salivary chromo- 
somes, two synapsed heterochromatic re- 
gions may be pulled apart, but show evi- 
dence of synapsis because they are still 
connected by strands of apparently sticky 
material. The right end of the fourth 
chromosome polynemes in Figure 1 1-6 
shows such glutinous matter, probably 
indicating synapsis with the chromocenter. 
Heterochromatin is chromatic and is not to 
be identified with the regions between bands; 
interband regions do not seem to contain the 



Fculgcn-stainable material - and apparently 
are achromatic, as is the spindle. 

3. Allopolyploidy 

Ploidy can increase another way besides 
allopolyploidy. Two species can each con- 
tribute two or more genomes to form a third 
species which is called an allopolyploid. 
Cultivated wheat is an allopolyploid. As 
expected, allopolyploids often show a com- 
bination of characteristics of their different 
parent species. This type of polyploidy is 
discussed in more detail in Chapter 18. 

Changes in genome number represent the 
class of normal and mutational events in- 
volving the largest unit of genetic material. 
Although many plants are polyploid and one 
plant has 512 chromosomes, polyploidy will 
produce a chromosome number that is un- 
wieldy in nuclear division if it occurs many 
times in succession. It should also be noted 
that certain other classes of mutation, like 
those involving a single locus, have greater 
difficulty expressing themselves in autopoly- 
ploids than they have in haploids or diploids, 
in which no other, or just one other, homol- 
ogous locus is able to mask the mutant effect. 

Aneusomy 

The next category of mutations to be dis- 
cussed involves the addition or subtraction 
of part of a chromosome set. Such muta- 
tions upset the normal chromosomal and 
gene balance and produce aneuploid ("not 
right-fold") chromosomal (genetic) consti- 
tutions by having the incorrect number of 
particular chromosomes (aneusomy). By 
what mechanisms can single whole (un- 
broken) chromosomes be added to or sub- 
tracted from a genome? 

1 . In Drosophila 

Recall that nondisjunction in the germ 
line of Drosophila can produce offspring, 

- See D. M. Steffensen (1963). 



156 



CHAPTER 1 1 



otherwise diploid, that arc XO, XXX, and 
XXY. Nondisjunction of the small fourth 
chromosome can load to the production of 
individuals with one fourth chromosome 
(huplo-IY individuals) or three (triplo-IY 
individuals) (Figure 11-7) being in this 
respect monosomic and trisomic, respec- 
tivelj : instead of disomic as is normal. Even 
though addition or subtraction o\' a chromo- 
some IV makes visible phenotypic changes 
from the disomic condition as can be seen 
from the phenotypes, both aneusomic 
changes are viable. On the other hand, 
individuals monosomic or trisomic for cither 
of the two large autosomes die before com- 
pleting the egg stage. 

When triploid Drosophila females — with 
all chromosomes trisomic — undergo meiosis, 
bundles of three homologous chromosomes 
(trivalents) may be formed at synapsis. 
This is because, at one place along the length 
of a chromosome, the pairing is between two 
homologs, and at another place it is between 
one of these two and the third homolog. 
In this way, although pairing is two-by-two 



at all levels, all three homologs are held to- 
gether as a divalent. At the first mciotic 
division the two homologs that are synapscd 
at their ccntromeric regions, separate, and 
go u> opposite poles, while the third homolog 
goes to either one of the poles. At the end 
o\' the second mciotic division, two nuclei 
each have one homolog of the trivalent. and 
two nuclei each have two homologs. The 
same result is obtained when synapsis is 
entirely between two homologs and excludes 
the third. Since each of the four trisomies 
present at metaphase I segregates independ- 
ently, eggs are produced which have one of 
the following: 

1 . Each chromosome type singly and, 
therefore, contain one complete ge- 
nome (being haploid) 

2. Two chromosomes of each type and, 
therefore, contain two genomes (be- 
ing diploid) 

3. Any combination in which some chro- 
mosomes are represented once and 
others twice (being aneusomic). 



figure 11-7. Haplo-IV (left) and triplo-IV (right) females of D. melano- 
gaster. The haplo-IV is smaller than the wild-type female shown in Fig. 2-6. 
(Drawn by E. M. Wallace.) 




Changes Involving Unbroken Chromosomes 



157 



We see, therefore, that meiosis produces 
many aneusomic gametes when the number 
of homologs is odd, as it is in triploids, 
pentaploids, etc. In tetraploids, since each 
chromosome can have a partner at meiosis, 
the four homologs often segregate two and 
two. Sometimes, however, the four homo- 
logs form a trivalent and segregate three and 
one, so that some aneusomic gametes are 
produced by polyploids with even numbers 
of homologs. 

Because the phenotypic effect of any gene 
depends directly or indirectly upon the 
phenotypic effects of most, if not all, of 
the other genes present, it is expected that 
a diploid individual contains, in its two sets 
of chromosomes, a proper balance of genes 
for the production of a successful phenotype. 
It is not surprising, then, that a haploid in- 
dividual mated to a diploid produces very 
few progeny, since after fertilization most 
zygotes are chromosomally unbalanced by 
the absence of one or more chromosomes 
needed to make two complete genomes. 
Mated to a diploid the triploid individual 
also produces zygotes that are imbalanced 
but in the opposite direction, having one or 
more chromosomes in excess of two ge- 
nomes. 

In matings with diploids, however, the 
triploid individual usually produces more 
offspring than the haploid. This observa- 
tion can be explained as the result of the 
lesser imbalance brought about by the addi- 
tion of chromosomes to the diploid condi- 
tion than by the subtraction of chromosomes 
from it. This effect can be seen by com- 
paring how far from normality (diploidy) 
each of the two abnormal conditions is. 
When one chromosome is in excess, the ab- 
normal chromosome number of three is one 
and a half times larger than the normal num- 
ber of two; when one chromosome is miss- 
ing, the abnormal chromosome number of 
one is two times smaller than the normal 
number. Thus, the addition of a chromo- 



some makes for a less drastic change in bal- 
ance than the subtraction. Accordingly, 
knowing that the triple dose of a large auto- 
some is lethal in Drosophila, we can cor- 
rectly predict that the single dose is lethal 
also. In these cases, death is attributable 
to genetic imbalance due to an excess of the 
genes present in a long autosome in trisomic 
individuals and to a deficiency of these 
genes in monosomic individuals. 

2. In Datura 

Chromosome addition and subtraction can 
also be studied in Datura 3 whose haploid 
chromosome number is twelve. It is pos- 
sible to obtain twelve different kinds of in- 
dividuals, each having a different one of the 
twelve chromosomes in addition to the dip- 
loid number. Each of these trisomies is 
given a different name such as "Globe." It 
is also possible to obtain viable plants that 
are diploid but missing one chromosome of 
a pair; these are monosomies or haplosomics. 
Individuals with two extra chromosomes of 
the same type (tetrasomics) or with two 
extra chromosomes of different types (dou- 
ble trisomies) are also found. 

Datura enables us to test the phenotypic 
consequences of disturbing the normal bal- 
ance among chromosomes. Compare, in 
Figure 11-8, the seed capsules of the normal 
diploid (2N) with those of diploids having 
either one extra chromosome (2N + 1 ) of 
the type producing Globe or two of these 
(2N + 2). The latter two polysomics can 
be called trisomic diploid and tetrasomic 
diploid, respectively. Although the tetra- 
somic is more stable chromosomally (each 
chromosome can have a partner at meiosis) 
than is the trisomic, the tetrasomic pheno- 
type is too abnormal to establish a race, 
since it has a still greater genetic imbalance 
than the trisomic and produces a still greater 
deviation from the normal diploid phenotype. 

3 Based upon work of A. F. Blakeslee and J 
Belling. 



15S 



( ii \r i i i' II 




t 



TETRAPLOID AN 



#• ##? 



2N I 2N 2 

IGlobel 



4N I 4N 2 4N 3 



FIGUR] 11-8. /.//t'<7 upon f/ie capsule of 
Datura of the presence of one or more extra 
"Globe" chromosomes. 



In comparison, the tetraploid (4N) indi- 
vidual is phenotypically almost like the dip- 
loid, since chromosomal balance is undis- 
turbed. The tetraploid which has one extra 
Globe chromosome (4N -+- 1, making it a 
pentasomic tetraploid) deviates from the 
tetraploid in the same direction as the 
2N -f- 1 deviates from 2N, but does so less 
extremely. Hexasomic tetraploids (4N + 2) 
de\ iate from 4N just about as much as 
2N + 1 deviates from 2N. It is clear, 
therefore, that adding a single chromosome 
to a tetraploid has less phenotypic effect 
than its addition to a diploid, since the shift 
in balance between chromosomes is relatively 
smaller in the former than in the latter. 
Thus, polyploids can stand whole chromo- 
some additions and subtractions better than 
diploids can. 

Since crosses between tetraploid Datura 
produce fertile seed in amounts sufficient to 
maintain a tetraploid race, the question 
arises, can a tetraploid race of Drosophila 
be produced? As mentioned, the gametes 
of a tetraploid Drosophila female contain 
complete genomes more often than do those 
of triploids. Since it produces many diploid 
eggs, the tetraploid female presents no diffi- 
culty for the continuity of a tetraploid race. 



To be o\' normal sex, a tetraploid male has 
to carry 2X + 2Y + 4 sets of A (Chapter 
8). But the X's (and Y's) in such a male 
usually synapse with each other during meio- 
sis so that after meiosis each sperm carries 
IX and I Y in addition to the 2A sets. In 
fertilizing eggs (from tetraploid females) 
containing 2X + 2 sets of A. sperm of this 
type produce zygotes with 3X -f- 1Y + 4 
sets of A which develop as sterile inter- 
sexes. Thus, a self-maintaining tetraploid 
race of Drosophila cannot be established. 
In fact, we can conclude that any species 
containing a heteromorphic pair of sex chro- 
mosomes (such as X and Y) cannot form 
polyploid races, since the correct balance 
between sex chromosomes and autosomes is 
upset by the meiotic divisions. This factor 
probably explains why polyploid races and 
species are rarer among animals than among 
plants whose sexuality (as in monoecious 
forms) is not associated with heteromorphic 
homologs. 

3. /// Man 

Down's syndrome, or mongolism, in hu- 
man beings is sometimes the result of a 
trisomic diploid chromosomal constitution. 
In this case, the trisomic is number 21 the 
third smallest of all human chromosomes 
(the smallest being the Y) (Figures 11-9, 
1 1-10). Trisomies for several other of the 
smaller autosomes are also known, each pro- 
ducing its own characteristic set of congenital 
abnormalities. Trisomy for the largest auto- 
somes is apparently lethal before birth, prob- 
ably due to the imbalance of too many 
genes. The very severe phenotypic defects 
observed among the least affected auto- 
somals trisomic individuals makes it a rea- 
sonable expectation that the monosomic 
condition of any autosome is lethal before 
birth — in accordance with the view that 
chromosome subtraction is even more detri- 
mental than chromosome addition. 



Changes Involving Unbroken Chromosomes 



159 




figure 1 1-9. The chromosomal complement of a normal human female. 
Cell was in mitotic metaphase {hence chromosomes appear double except at 
the centromere) when squashed and photographed. (Courtesy of T. C. Hsu.) 



The frequency among live births of 
Down's syndrome due to trisomy has been 
determined as approximately 0.2%. Most 
cases of Down's syndrome occur among the 
children of older mothers and are due pri- 
marily to nondisjunction during oogenesis. 
If other chromosomes have a similar fre- 
quency of nondisjunction, there might be a 
minimum of 4.4% (22 X 0.2% ) of zygotes 
autosomally trisomic at conception. There 
might also be another 4.4% of zygotes that 
are autosomal monosomies, due to the equal 
chance that the haploid meiotic product 
complementary to the one which is disomic 



— the nullosomic one — becomes the egg. In 
fact, more nullosomic than disomic gametes 
are expected, since a chromosome left out 
of one daughter nucleus need not be included 
in the sister nucleus. Supporting a normally 
high frequency of aneusomy is the observa- 
tion that about one quarter of aborted hu- 
man fetuses show a chromosomal derange- 
ment. It is expected, moreover, that main 
conceptions involving aneusomy. especially 
monosomy, are lost so early in pregnanes 
that they go unnoticed. 

Nondisjunction leading to aneusomy can 
also occur in the paternal germ line of man, 



160 



CHAPTF.R I I 



1) 11 (1 

1 2 3 


I) H n 

4 5 6 


II 

X 


1) if II 

7 8 9 


11 If 

10 11 


II 

12 


ft it tt 

13 14 15 


16 17 


li 

18 


** I » 

id 20 


A 4 A 

21 


II 

22 



figure 11-10. Chromosomal constitution found in a female showing Down's syndrome. 
(By permission of M. A. Ferguson-Smith and A. W. Johnston, and The Annals of 
Internal Medicine, vol. 53, p. 361, I960.) After photographing a squash preparation 
like that in Figure 11-10, the chromosomes are cut out and "paired" as shown here. 



although this does not seem to contribute 
very significantly to the total observed fre- 
quency. In man, almost all nondisjunction 
is associated with the aging of oocytes. 
The reverse is true in the mouse, however, 
even though mouse females — like human fe- 
males — are born with all their germ cells 
in the oocyte stage. Thus, in the mouse, 
marked chromosomes show that spontaneous 
aneusomy almost always has a paternal 
origin. It should be noted that, in the 
mouse, viable aneusomy also occurs for the 
sex chromosomes and for certain small auto- 
somes when trisomic.' 
4 See A. B. Griffen and M- C Bunker (1964). 



The incidence of nondisjunction can be 
increased by high energy radiations. Carbon 
dioxide, other chemical substances, and cer- 
tain diploid genotypes can increase the non- 
disjunction rate in Drosophila. In human 
beings, the evidence that older women are 
more apt to have trisomic children suggests 
that some metabolic defect associated with 
increased age increases the chance for non- 
disjunction. 

Although chromosome loss may result 
from spontaneous mciotic and nonmeiotic 
nondisjunction in diploids as well as from 
the meiotic process normally taking place 
in polyploids, it should not be inferred that 



Changes Involving Unbroken Chromosomes 



161 



these are the only ways entire chromosomes 
can be lost. 

Mosaic Heteroploidy and Aneusomy 

Mutations leading to heteroploidy need not 
involve germ cells or the entire organism, 
as mentioned with respect to asexually re- 
producing species (p. 152). Sexually repro- 
ducing species of plants and animals may 
also show mosaicism for ploidy involving 
reproductive or nonreproductive tissues, or 
both. In man, for example, a baby boy 
has been studied who is diploid in some 
tissues and triploid in other, normally dip- 
loid, tissues. About 3% of cells in certain 
human tissue cultures show such changes in 
ploidy. 

Aneusomy can also originate at any 
mitotic, as well as meiotic, nuclear division. 
Thus, nondisjunction at the first nuclear di- 
vision of a normal human zygote might pro- 
duce one nucleus that is monosomic and one 
trisomic for chromosome 21. In this case 
the former nucleus is expected to die, and 
the latter nucleus, to produce a completely 
mongoloid individual. 

Some of the aneusomics born of older 



mothers may have originated in such a post- 
zygotic nondisjunction, as is the case in 
mice. If nondisjunction occurs later in 
development, it produces complementary 
monosomic and trisomic mutant patches in 
a diploid background, which — in the case of 
autosomes in man and mouse — are usually 
expected to be lethal to the individual. That 
such nondisjunctions or chromosome losses 
do occur with appreciable frequency is sug- 
gested by the frequent occurrence in human 
adults of a few cells per hundred which are 
scored as having one or two chromosomes 
too few or too much. It is extremely un- 
likely that all, or even most, of these abnor- 
mal counts are due to experimental errors 
in preparing or in scoring the cells. Under 
normal circumstances one would expect the 
aneusomic cells produced after birth to be 
functionally inferior to their neighboring 
euploid cells and, therefore, at a selective 
disadvantage. 

Because of the large genetic unbalance it 
produces, addition and subtraction of whole 
chromosomes is a class of mutation which 
involves a phenotypic change too drastic to 
play a very significant role in evolution. 



SUMMARY AND CONCLUSIONS 

The mutational events involving the largest recombinational unit of genetic material 
are euploid changes in the number of whole sets of chromosomes — heteroploidy. Ploidy 
can increase by allopolyploidy, autopolyploidy, and polynemy. The modes of origin 
and the breeding behavior of autopolyploids, and the origin and structure of the giant 
polynemic chromosomes in the salivary gland of Drosophila larvae are considered in 
detail. 

Loss or gain of part of a genome — aneuploidy — can result from nondisjunction and 
the segregation of chromosomes in polyploids, especially those possessing an odd num- 
ber of genomes. Not only do such mutations occur in the germ and somatic lines 
spontaneously, but they may be initiated or have their frequency enhanced by physical 
and chemical factors. 

The addition or subtraction of single chromosomes results in aneusomy. The ab- 
sence of a chromosome is more detrimental to survival than an excess. Aneusomy 
produces too drastic a phenotypic change to be as inportant in evolution as heteroploidy. 



Ki2 CHAPTER 1] 

REFERENCES 

^uerbach, C, Mutation. An Introduction to Research on Mutagenesis. Part I. Meth- 
ods, Edinburgh: Oliver and Boyd, 1962. 

Blakeslee, \. I .. "New Jimson Weeds from Old Chromosomes," J. Hered., 25:80 108, 
1934. 

Blakeslee, \. I . and Belling, J., "Chromosomal Mutations in the Jimson Weed. Datura 
Stramonium," J. Hered., 15:194 206, l l >24. 

Bridges, C. G., and Brehme, K. S., The Miliums oj Drosophila Melanogaster, Wash- 
ington. D.C.: Carnegie Institution of Washington, Publ. 552. 1944. 

Burdette, W. J. (Ed.). Methodology in Mammalian Genetics, San Francisco: Holden- 
Day, Inc.. 1963. 

Dobzhansky, lh.. Genetics and the Origin of Species, 2nd Ed.. New York: Columbia 
University Press. Chap. 7. pp. 223-253. 1941. 

Griffen, A. B., and Bunker. M. C. "Three Cases of Trisomy in the Mouse." Proc. Nat. 
Acad. Sci.. U.S.. 52:1194-1198. 1964. 

Heitz. E.. and Bauer. H., "Beweise fiir die Chromosomennatur der Kernschleifen in 
den Knauelkernen von Bibio hortulanus L. (Cytologische Untersuchungen an 
Dipteren, 1)." Z. Zellforsch.. 17:67-82. 1933. 

Painter. T. S.. "A New Method for the Study of Chromosome Rearrangements and 
Plotting of Chromosome Maps," Science, 78:585-586, 1933. Reprinted in Classic 
Papers in Genetics, Peters. J. A. (Ed.), Englewood Cliffs, N.J.: Prentice-Hall, 
pp. 161-163. 1959. 

Patau, K.. Smith. D. W.. Therman, E., Inhorn. S. L., and Wagner, H. P., "Multiple 
Congenital Anomaly Caused by an Extra Autosome." Lancet, 1:790-793, 1960. 

Russell. L. B.. "Chromosome Aberrations in Experimental Mammals," Progress in 
Medical Genetics. 2:230-294. 1962. 

Steffensen, D. M.. "Evidence for the Apparent Absence of DNA in the Interbands of 
Drosophila Salivary Chromosomes." Genetics, 48:1289-1301, 1963. 

Suomalainen, E.. "Significance of Parthenogenesis in the Evolution of Insects," Ann. 

Rev. Ent., 7:349-366, 1962. 
White. M. J. D.. Animal Cytology and Evolution, 2nd Ed.. Cambridge: Cambridge 

University Press. 1954. 



QUESTIONS FOR DISCUSSION 

11.1. How do we know that the genetic differences in a population today were not 
always present? 

11.2. What have you learned in this chapter about the characteristics of mutation? 

11.3. What is the relation between mutants and genes? Mutants and recombination? 

11.4. From your present knowledge, how would you modify the statements on page 
1 1 relative to the ploidy of gametes? 

11.5. Describe at least two different ways that the trisomy causing Down's syndrome 

may originate. 

1 1.6. The only presently known case of trisomy for a chromosome of the 19-20 group 
occurred mosaically in a six-year-old boy. To what do you attribute this? 



Changes Involving Unbroken Chromosomes 163 

1 1.7. Discuss the statement: All somatic cells from diploid zygotes are chromosom- 
ally identical. 

I 1.8. Do you suppose that the human species will henefit from a discovery that certain 
of its members are trisomic? Explain. 

1 1 .9. What are the advantages of allopolyploidy? Of allopolyploidy? 

11.10. What genetic explanation can you offer for the fact, demonstrated in Figure 
1 1-2. that the seed capsule of the Datura haploid is smaller than that of the 
triploid? 

1.11. What do you consider to be the advantages and disadvantages of polynemy? 

1.12. Unfertilized mammalian eggs can contain ploidies of IN. 2N. 3N, or 4N. 
Explain how each of these could be produced. 

1.13. How can you explain the fact that persons with Down's syndrome are more 
susceptible to leukemia than normal diploids? 

1.14. Explain why individuals with Down's syndrome show a wide variety of pheno- 
typic differences as well as similarities in their abnormalities. 

1.15. Some babies classified as normal at birth are clearly mongoloid when a year 
old. What would you do to assure an early diagnosis of Down's syndrome? 

1.16. Would you expect a correlation between producing a child with Down's syn- 
drome and the frequency with which the mother has abortions? Subsequent 
children with Down's syndrome? Explain. 

1.17. Should a woman with a trisomic mongoloid sibling be more than ordinarily con- 
cerned about having a child of this type? Explain. 

1.18. After examining Figures 11-8 and 11-9, discuss the precision with which a 
given human chromosome can be identified. 

1.19. How would you proceed to determine the somatic chromosome composition 
of a given human individual? 

1.20. Discuss the phenotypic effects of adding an N-l genome to individuals that are 
normally N. 2N, 3N. or 4N. 

1.21. R. A. Turpin reported two cases of monozygotic twins. One set contains an 
XY male and an XO female; the other set is composed of a disomic-21 male 
and a trisomic-21 male. Discuss the mechanisms probably involved in pro- 
ducing such twins. Include in your hypothesis the additional fact that one XO 
cell is also found in the first XY individual mentioned. 



Chapter 12 

STRUCTURAL CHANGES 
IN CHROMOSOMES 



T 



|he two classes of mutation 
dealt with in Chapter 11 in- 
volved changes in chromo- 
somal content of unbroken individual or sets 
o( chromosomes. In some instances, mu- 
tants are based upon the gain, loss, or shift 
of a part of one or more chromosomes. 
All such structural changes are preceded 
by chromosome breakage, which — ignoring 
chromatids for the present — results in two 
new. "sticky" ends. When several breaks 
are produced, the new ends can join together 
but only in pairs, any new end capable of 
joining any other new end. Moreover, an 
end produced by breakage cannot join the 
normal (unbroken) end of a chromosome. 
Thus, originally free ends of chromosomes 
are not sticky because they have genes, 
called telomeres, which serve to seal them 
off, making it impossible for a normal end 
to join to any other. 

The two ends produced by one break 
usually join together in what is called resti- 
tutional union even when ends produced by 
other breaks coexist in the same nucleus. 
This indicates that proximity favors the 
union of sticky ends. Although restitu- 
tional union usually occurs and thereby 
restores the original linear order of the chro- 
mosome, the ends uniting may sometimes 
come from different breaks, so that a new 
chromosomal (gene) arrangement is pro- 
duced. The latter union is, therefore, a non- 
restitutioncd, or exchange, or cross-union 
type. Let us see how nonrestitutional unions 
164 



produce various structural changes in chro- 
mosomes. 

Consequences of a Single Chromosome Break 

Consider first the consequences of a single 
chromosome break; that is. a break through 
both chromatids (Figure 12-1). Diagram 
1 represents a norma] chromosome (its chro- 
matids are not shown), whose centromere 
is indicated by a black dot. In diagram 2 
this chromosome is broken. If the new 
chromosome ends join together, that is. 
restitute, no chromosomal rearrangement is 
produced. Although restitution usually oc- 
curs, it may sometimes fail because the new 
ends spring apart or are moved apart by 
Brownian movement or protoplasmic cur- 
rents. In nonrestitution. chromosome repli- 
cation produces a daughter chromosome just 
like the parent — with a break in the same 
position — as shown in diagram 3 where the 
two broken sister chromosomes are indi- 
cated. The union of the piece containing 
no centromere ( a ) to the centromere-bear- 
ing piece of the other sister chromosome 
( b' ) would, in effect, be restitution as would 
the joining of a' to b. (Sometimes, only 
one of these restitutional unions occurs.) 

If restitution does not occur before or 
after the chromosome replicates, the ends 
closest together usually join together, these 
being the corresponding ends of the sister 
chromosomes (a with a' and b with b'). 
As shown in diagram 4. the results of such 
nonrestitutional unions are one chromosome 
with no centromere (an acentric chromo- 
some), and one with two (a dicentric chro- 
mosome). Note that both the acentric and 
the dicentric chromosomes are composed of 
identical halves lengthwise, each, therefore, 
being termed an isochromosome. (This dia- 
gram shows the chromosomes contracting 
preparatory to metaphase.) 

In diagram 5 we can see that in mitotic 
anaphase the acentric isochromosome is not 
pulled toward either pole, whereas the dicen- 



Structural Changes in Chromosomes 



165 



trie one is pulled toward both poles at once. 
The acentric isochromosome is, therefore, 
not included in either daughter nucleus and 
so is lost to both. (The acentric pieces in 
diagram 3 will be lost in any subsequent 
division, whether they are joined to each 
other or do not join at all.) The dicentric 
isochromosome. being pulled to both poles 
at once, forms a bridge. A bridge can pre- 
vent any part of the chromosome from enter- 
ing either daughter nucleus, so that the 
dicentric is lost. Alternatively, the centric 
regions of the dicentric piece can enter the 
daughter nuclei, and the bridge can either 
snap at any one of a number of places be- 
tween the centromeres and free the daugh- 
ter nuclei from each other, or it can persist, 
joining the daughter nuclei together. 

The amount of phenotypic detriment that 
a single nonrestituting chromosome break 
will produce in the daughter cells and their 
progeny cells depends upon the particular 
chromosome involved, the place of breakage, 
and the fate of the dicentric piece. Suppose. 
for example, that chromosome IV of Dro- 
sophila (often viable as a haplo-IV indi- 
vidual) is the chromosome involved. The 
break can occur at any position in IV, and 
the loss of the genes in the acentric piece, 
though detrimental, does not usually cause 



death; neither does the loss of the entire 
dicentric fragment if excluded from both 
daughter nuclei, nor, probably, does a snap 
in the bridge between the daughter nuclei. 
(In the last case, each daughter nucleus is 
deficient, at least for the genes in the acen- 
tric piece.) 

Note what happens when a bridge, involv- 
ing a dicentric isochromosome linearly dif- 
ferentiated as a.bcddcb.a (the centromere 
is between a and b). does not snap between 
the d's. If it breaks between b and c, one 
fragment is even more deficient (yet viable 
in this example), whereas the other con- 
tains an extra dose of the genes in the cd 
region (and is most probably viable). Re- 
gardless of where the bridge snaps, both 
daughter nuclei carry a centric fragment 
which, after replicating, usually forms a new 
dicentric isochromosome and can again form 
a bridge at the next mitotic division. It is 
possible, therefore, to have bridge-breakage- 
fusion-bridge cycles in successive nuclear 
generations. 

When a bridge fails to break leaving the 
two daughter nuclei tied together, the en- 
tanglement of the nuclei may interfere with 
subsequent attempts at nuclear division. In 
our example, this interference may be of 
much greater importance than the presence 



figure 12-1. Conse- 
quences of a single 
nonrestituting 
chromosome break. 



w 



> < 




!(>(• 



( ii \i- ni< I 2 



or absence of all of the genes located in the 
bridge. 

Suppose, however, that the broken chro- 
mosome is one of the largo autosomes of 
Drosophila. Detriment or death to one or 
both daughter cells may occur because of 
the genes lost when either the acentric or 
the dicentric fragment is left out of one 
or both daughter nuclei. In addition, suc- 
cessive bridge-breakage-fusion-bridge cycles 
ma) harm future cell generations via the 
abnormal quantities o\' chromosomal regions; 
that is. the ancuploidy. resulting from the 
off-center breakage of dicentric isochromo- 
somes. Other things being equal, shorter 
dicentrics are expected to break more often 
than longer ones. Of course, any inter- 
nuclear bridge that does not break may frus- 
trate future nuclear division. 

Single chromosome breaks can occur in 
either the somatic or the germ line. In the 
latter case, aneuploid gametes may be pro- 
duced. Since the genes are found to be 
physiologically inactive in the gametes of ani- 
mals, aneuploid genomes can enter the egg 
and sperm without impairing their function- 
ing (as implied on p. 104). Accordingly, 
in animals, aneuploid genomes can be carried 
by unaffected gametes into the zygote, which 
may subsequently suffer dominant harmful 
or lethal effects. In many plants, however, 
the products of meiosis form a gametophyte 
generation which performs physiological 
functions requiring gene action, in which 
case, ancuploidy is usually more lethal or 
detrimental before fertilization than after. 

Chromatid Breaks 

A break can be produced in one and not the 
other chromatid of a chromosome. Such 
chromatid breaks are more likely to restitute 
than chromosome breaks, since the unbroken 
strand serves as a splint to hold the newly- 
produced ends close to each other. What 
appears under the microscope as a break 
involving only one chromatid may initially 



have been a chromosome (or isochromatid) 

break that was followed by restitution of one 
but not (yet) the other chromatid. 

Nonrestituted chromatid fragments be- 
come nonrestituted chromosome fragments 
if they persist long enough to replicate. To 
be seen cytologically, a conjoined chromatid 
or chromosome break produced during inter- 
phase usually has to persist until nuclear 
division occurs. Some chromatid and. per- 
haps, chromosome breaks induced in con- 
tracted ( metaphasc ) chromosomes may not 
be visible, the pieces being held together 
without joining by the nongenetic auxiliary 
material in a chromosome. To detect such 
unjoined breaks one would have to wait until 
the next division. Essentially all ends pro- 
duced by breaks are not sticky when the 
chromosome is contracted as during nuclear 
division; joinings are restricted largely, if not 
completely, to the period between late telo- 
phase and early prophase. Accordingly, the 
later in this period a break is produced, the 
less likely it is that the ends will join; broken 
ends produced between early prophase and 
late telophase have the maximum time for 
joining but probably also the maximum op- 
portunity to cross-unite. 

For simplicity, the discussion which fol- 
lows is restricted to isochromatid breaks that 
fail to restitute. The reader is given the 
task of working out the consequences of 
aneuploidy resulting from single nonresti- 
tuted broken chromatids. The lack of fur- 
ther discussion on this type of mutation does 
not reflect on the relative frequency or im- 
portance of chromosome versus chromatid 
breaks. Agents capable of producing chro- 
mosome breaks can also produce chromatid 
breaks; moreover, certain agents may prefer- 
entially produce chromatid breaks. 

Consequences of Two Breaks in One 
Chromosome 

When a chromosome is broken twice, the 
two breaking points may be paracentric, that 



Structural Changes in Chromosomes 



167 



PARACENTRIC BREAKS 
ABCDEFGHIJ 



r 

A G H I J Deficiency 

C D 
B C D E F or B C J E lost 

F 

(a) 



AFEDCBG HIJ 

Inversion 
(b) 



PERICENTRIC BREAKS 
ABCDEFGHIJ 



A B C D I J lost 



J Deficiency 



G F 



ABCDHGFEIJ 



Inversion 



(c) (d) 

FIGURE 12—2. Some consequences of two breaks in the same chromosome. 



is, to one side of the centromere, or peri- 
centric, that is, with the centromere between 
them (Figure 12-2). 

1. Deficiency 

Consider a chromosome linearly differen- 
tiated as ABCDEFG.HIJ, the centromere 
being between G and H. When the breaks 
are paracentric in position (for example, 
between A and B, and between F and G). 
the fragments can unite to produce a centric 
chromosome (AG. HIJ, Figure 12-2a) de- 
ficient for the acentric interstitial (nonter- 
minal) piece (BCDEF). The ends of the 
latter fragment may join to produce a ring 



chromosome, or they may not. In either 
event, the acentric fragment is usually lost 
before the next nuclear division. When the 
breaks are pericentric (for example, between 
D and E, and between H and I ), the acentric 
end pieces are lost, even if they join together 
(Figure 12-2c). The middle centric piece 
can survive if its ends join to form a ring 
and if the deficient sections are not exten- 
sive. Even if a ring survives because it is 
not too hypoploid (the aneuploid condition 
in which genes or chromosomal regions are 
missing), it is still at a disadvantage because 
a single crossing over either with a nonring 
(rod) homolog or with another ring results 



L68 



CHAPTER 12 



in a dicentric rod or ring, respectively, as 
can be seen by drawing the appropriate con- 
figurations. 

Of course, a nondividing nucleus, in which 
breakage or another structural change oc- 
curs, is still euploid. The first occurrence of 
hypoploid)/ or hyperploidy (aneuploidy due 
to an excess of genes or chromosome parts) 
is in the daughter nuclei formed by such a 
nucleus. This delay in producing an aneu- 
ploid nucleus should be remembered when 
we state that chromosomes with small de- 
ficiencies can be lethal when homozygous, 
and detrimental when heterozygous; chro- 
mosomes with large deficiencies usually act 
as dominant lethals in the next cell genera- 
tion. Remember also that we have ignored 
— and shall continue to do so for the rest of 
the chapter — the usual consequence of two 
breaks, that is. restitution for all ends pro- 
duced by breakage. 

2. Inversion 

Another structural consequence of two 
breaks in the same chromosome is repre- 
sented in Figure 1 2-2 b and d. In this case, 
the middle piece is inverted with respect to 
the end pieces and undergoes exchange 
unions with them. The result which is either 
a paracentric or a pericentric inversion (Fig- 
ure 12-2 b and d), is due to the middle seg- 
ment moving while the ends are relatively 
stationary, or the reverse. Note that inver- 
sion is a euploid rearrangement. 

Structural rearrangements in chromo- 
somes can occur in either the somatic or 
the germ line. An inversion which occurs 
in the germ line may be retained in the pop- 
ulation long enough to become homozygous 
in some individuals. Meiotic behavior is 
normal in such inversion homozygotes 
whether or not the tetrad undergoes cross- 
ing over, since all the strands in the tetrad 
are identically inverted. Other individuals 
in the population, however, may possess one 
inverted and one noninverted homolog, be- 



ing inversion heterozygotes. Provided the 
inversion is wry small, these homologs will 
pair properly everywhere but in the inverted 
region. Because the homologs cannot twist 
enough to make homologous loci meet in so 
short a region, they will fail to synapse and 
no crossing over will occur. Insofar as 
crossing over can lead to more adaptive re- 
combinants, such inversion heterozygotes are 
at a disadvantage compared to noninversion 
or inversion homozygotes because of the ab- 
sence of recombination among genes within 
the inverted region. Nevertheless, very 
small inversions do survive in many species. 

Consider the meiotic process in heterozy- 
gotes for larger paracentric inversions. In 
this case (Figure 12-3A), synapsis between 
homologs occurs for all regions except those 
adjacent to the points of breakage. This 
synapsis requires one homolog twisting in 
the inverted region while the other does not. 
The figure happens to show the inverted and 
not the noninverted chromosome twisting, 
but the reverse is equally likely to occur. 
If crossing over occurs anywhere outside the 
inverted region, each of the four meiotic 
products will be eucentric (having one cen- 
tromere), as usual. If, however, a single 
crossing over occurs anywhere within the 
region inverted — as shown between C and 
D — the two noncrossover strands of the 
tetrad will be eucentric (one with and one 
without the inversion), and the two cross- 
overs will be aneucentric (having more than 
one centromere or none). One of the aneu- 
centrics will be acentric (duplicated for A 
and deficient for G.HIJ); the other will be 
dicentric (deficient and duplicated for these 
respective regions). If the inversion is only 
moderately long, only one crossing over can 
occur within it; if sufficiently long, double 
crossing over is possible. When such double 
crossing over is of the two-strand type, both 
crossover strands are eucentric. 

In animals, gametes function regardless 
of the ploidy of the meiotic products they 



Structural Changes in Chromosomes 



169 




A. PARACENTRIC INVERSION 



iA 



.A«B"C — D 










B. PERICENTRIC INVERSION 

figure 12-3. A single crossing over in an 
inversion heterozygote. {See text for explana- 
tion.) 



contain. If a gamete contains an aneucentric 
produced by crossing over in a paracentric 
inversion heterozygote, this chromosome will 
usually have a dominant lethal effect after 
fertilization; that is, individuals heterozygous 
for moderate to large paracentric inversions 
are at reproductive disadvantage, which often 
leads to the elimination of the inversion from 
the population soon after it arises by muta- 
tion. This disadvantage is partly avoided in 
those species having no crossing over in one 
sex. For example, in the Drosophila male 
each homolog, inverted or not, is a noncross- 



over and has the same chance of being in- 
cluded in the gamete. A special factor oper- 
ates during meiosis in some species in which 
the female undergoes crossing over, occur- 
ring only if the two meiotic divisions occur 
in tandem, as they do in female Drosophila. 
In the Drosophila oocyte heterozygous for 
a paracentric inversion, a single crossing 
over within the inverted region produces the 
usual dicentric at anaphase I. But this di- 
centric serves to hold the dyads at metaphase 
II, so that the two eucentric monads proceed 
to the two outermost of the four poles. 
Therefore, at the end of telophase II, the cen- 
tric meiotic products are arranged in a row: 
eucentric; part of dicentric; remainder of di- 
centric; eucentric. It is one of the two end 
eucentric-containing nuclei which becomes 
the egg nucleus, the others degenerate. In 
this way the dicentric strand is prevented 
from entering the nucleus that becomes 
gametic; the gamete, therefore, receives one 
of the two eucentric, noncrossover strands. 
That is why in Drosophila, paracentric in- 
versions of any size rarely cause aneuploid 
gametes in either sex and can become estab- 
lished in nature. 

What products result from a crossing over 
within the inverted region in a heterozygote 
for a larger pericentric inversion? As seen 
in Figure 12-3B, a single crossing over, such 
as between F and G, produces four eucentric 
strands: two noncrossovers (one with and 
one without the inversion) ; one with a dupli- 
cation (for ABCD) and a deficiency (for 
IJ) ; the last with a deficiency and a duplica- 
tion of the respective regions. All strands 
enter the gametes of males if crossing over 
occurs in the male. Each strand also has 
an equal chance of being present in the 
gametes of females capable of crossing over. 
This is true even in Drosophila where shunt- 
ing of euploid strands into the egg nucleus 
does not occur because all the meiotic prod- 
ucts are eucentric. Consequently, aneuploidy 
which results from crossing over within a 



170 



( II \PTHR 12 



BREAKAGE 



ABCDEFG HIJ 



REPLICATION 



A B C D E F G HIJ 



AB CDE FGHIJ 



i u.i ri I 2 4 ( right ) . 
Duplication ( tandem 
type ) . 



CROSS-UNION 



ABCDECDE FG HIJ 



A B FGHIJ 



FIGURE 12-5 (below). Reciprocal transloca- 
tion between nonhomologous chromosomes. 

K L M N O 



P Q R S T U 



ANEUCENTRIC TYPE: 
K L M N O 



Q R S T U 



EUCENTRIC TYPE: 
K L Q R S T U 



P M N O 



pericentric inversion always puts the hetero- 
zygote at a reproductive disadvantage. For 
this reason, only the smallest pericentric in- 
versions — those which do not synapse when 
heterozygous — are usually able to survive in 
the wild. 

3. Duplication 

If, following two breaks in the same chro- 
mosome, joining is delayed until after the 
broken chromosome reproduces ( Figure 
12-4). the two interstitial pieces and the 
appropriate end pieces can join to produce 
a eutelomcric chromosome with the inter- 
stitial region repeated. This rearrangement 
is called a duplication. Neither, either, or 
both of the regions involved in the duplica- 
tion can be inverted with respect to the orig- 
inal arrangement. The two remaining end 
pieces may or may not join to form a de- 
ficient chromosome. (A deficiency can also 
be produced without a duplication when the 
end pieces join before chromosome replica- 
tion.) Provided that the duplicated region 
is small enough and acentric, it can survive 
in nature. 



Structural Changes in Chromosomes 



171 



Consequences of Two Breaks in Two 
Chromosomes 

What happens when two breaks occur, one 
in each of two different chromosomes? In 
the first such case, the two broken chromo- 
somes are nonhomologous (Figure 12-5). 
If the two centric pieces unite, a dicentric 
is formed and the two acentric pieces are 
lost in the next division, whether they join 
each other or do not join at all. If all pieces 
join as indicated, then there is a mutual ex- 
change of segments between nonhomologous 
chromosomes, which is called a segmental 
interchange, or more often, a reciprocal 
translocation. This is the aneucentric type 
of reciprocal translocation and often acts as 
a dominant lethal in a subsequent division, 



particularly when the dicentric is pulled to- 
ward both poles at once. 

The reverse is often just as likely, how- 
ever; union occurs between the centric piece 
o\' one chromosome and the acentric piece 
of the nonhomolog, with the centric piece of 
the second joining the acentric piece of the 
first. This reciprocal translocation is of the 
dicentric type. In individuals heterozygous 
for such an exchange (Figure 12-6), hav- 
ing two nonhomologs translocated and two 
nontranslocated, gametes are formed with 
deficiencies and duplications if, by segrega- 
tion, they receive one but not both members 
of the reciprocal translocation. 

When the chromosomes in nuclei are com- 
pressed in a relatively small volume, no 
broken end is far from any other; usually, 



FIGURE 12-6. Diagrammatic representation of segregation in dicentric reciprocal translocation 
heterozygotes. (Chromatids not shown; the spindles — also not shown — have their poles oriented 
vertically.) 



<><» 



SYNAPSIS 



F 



Zigzag Circle 



/ \ 



Open Circle 



DIAKINESIS 

AND 

METAPHASE I 





TELOPHASE I 



GAMETES 



ViV 



All Euploid 



vv vv 

All Aneuploid (Half-Translocationall 



172 



CHAPTER 12 



if one of the two unions deeded tor recip- 
rocal translocation occurs, so does the other. 
Such is the case in the nucleus of the Dro- 
sophila sperm just alter fertilization. In 

oocytes and probably in other cells that have 
a relatively large nuclear volume, the dis- 
tance between the broken ends of nonhomo- 
logs is so great that reciprocal translocations 
are comparatively rare and. even if one cross 
union occurs, the two other broken ends 
usually tail to join to each other, so that 
onlj half of a reciprocal translocation — a 
half-translocation — is produced. The loss 
or behavior of the unjoined fragments usually 
causes descendent cells to die or to be ab- 
normal, as would be expected. Half-trans- 
locations can also result when heterozygotes 
for a eucentric reciprocal translocation un- 
dergo segregation (see Figure 12-6), and 
only one of the two reciprocals is present 
in a gamete. 

Some children with Down's syndrome 
have 46 chromosomes. These chromosomes 
include — in addition to two normal num- 
ber 21*s — an autosomal pair (from group 
13-15 or from group 16-18) which is het- 
eromorphic. one member being longer than 
usual. The extra piece is probably the long 
arm of 21, so that the individual is hyper- 
ploid for 21, being almost trisomic 21. In 
some cases the mother is phenotypically nor- 
mal although she is heterozygous for a 
eucentric reciprocal translocation between 
21 and, for example, 15. Her chromosome 
constitution can be represented by 15, 15.21 
(centromere of 15), 21.15 (centromere of 
21 ). 21. An egg containing 21 and 15.21 
(the half-translocation) fertilized by a nor- 
mal sperm (containing 21 and 15) produces 
the almost-trisomic-21 mongoloid under dis- 
cussion. (The break in 15 must have been 
so close to the end that the hypoploid seg- 
ment in the half-translocation mongoloid in- 
dividual was not lethal.) In other cases 
such half-translocational mongoloids have 
half-translocational nonmongoloid mothers 



with 45 chromosomes. These mothers have 
only one normal 21, one normal 15, for 
example, and the half-translocation 15.21. 
The hypoploidy for both 21 and 15 must 
be small enough to be viable in the mother, 
who can produce the aneuploid gamete that 
makes her child mongoloid. (Note in the 
cases cited above no relation exists between 
mother's age and the occurrence of half- 
translocational mongoloid children. ) 

In the second case in which two chromo- 
somes are broken once, the chromosomes 
are homologs (ABCDEFG.HIJ). The 
breaks are usually at different places, for 
example, between A and B in one chromo- 
some, and between D and E in the other. 
Here, also, reciprocal translocation can occur 
two ways. The aneucentric type produces 
a dicentric and an acentric chromosome 
whose fate can be readily predicted. The 
eucentric type produces two eucentric chro- 
mosomes, the BCD region being deficient 
in one and duplicated in the other. 

From the preceding discussion, one would 
expect eucentric reciprocal translocations to 
tend to be eliminated from the population 
soon after arising by mutation, since they 
are usually heterozygous and cause about 
50% of gametes to be half-translocational 
aneuploids. Certain eucentric reciprocal 
translocations, however, seem to be excep- 
tions. In these cases, almost a whole arm 
of each chromosome is mutually exchanged. 
Such whole-arm reciprocal translocations — 
when heterozygous in Drosophila and prob- 
ably in most other species — tend to synapse 
and disjoin in the following way: at synapsis 
the heterozygous reciprocal translocation 
forms an X configuration composed of two 
tetrads (Figure 12-6). Later, when homol- 
ogous centromeres repel each other, alter- 
nate centromeres move toward the same 
pole, so that as the chiasmata move towards 
the ends, a zigzag arrangement of four dyads 
results (Figure 12-6). Because of this al- 
ternate centromeric orientation, anaphase I 



Structural Changes in Chromosomes 



173 



produces one nucleus without the transloca- 
tion and the other with the full transloca- 
tion. Since euploid gametes are usually 
formed, such translocation heterozygotes are 
not at an appreciable reproductive disad- 
vantage. 

Increasing Gene Number 

Both here and in Chapter 1 I , it has been 
pointed out that a change in ploidy can sur- 
vive in nature when it involves either no 
shift in chromosome balance (because it 
deals with whole genomes) or eucentric 
aneuploidy due to small segments of chro- 
mosomes which are hypo- or hyperploid. 
In the latter cases, the number of deficient 
or duplicated genes is small enough to pro- 
duce a tolerable phenotypic effect. It is 
reasonable to assume that the greater the 
amount of chromosomal material, the greater 
the complexity possible in an organism and. 
consequently, the greater the diversity pos- 
sible in its phenotype and adaptiveness. Ac- 
cordingly, viable changes in ploidy must be 
particularly important in organic evolution. 
It is desirable, therefore, to specify some of 
the different ways that small numbers of 
genes can be added to a genome after break- 
age. 



Two methods of increasing gene number 
after breakage have already been described. 
One requires two breaks in the same chro- 
mosome; the entire chromosome then repli- 
cates, after which the broken ends join to 
form a chromosome with the interstitial 
piece duplicated (p. 170); the other involves 
each member of a pair of homologs break- 
ing once in a different region before eucen- 
tric cross union (p. 172). 

A third mechanism involves three breaks 
in one chromosome. The two interstitial 
pieces exchange positions, producing what 




v 



"> 



figure 12-7. Inversion het- 
erozygotes in corn (pachy- 
nema) (courtesy of D. T. 
Morgan, Jr.) and in Drosoph- 
ila (salivary gland) (courtesy 
of M. D enter ec). 




171 



CHAPTER 12 



i 



m 






^ 



v. 



# % 

v 



*„ 



t 









I [GURE I 2 8. Salivary gland 
chromosomes heterozygous for a 
shift within the right arm of chro- 
mosome 3 of Drosophila melano- 
gaster. A piece from map region 
"98" is inserted into map region 
"91." I lie rightmost buckle is 
clue to the absence of the shifted 
segment; the leftmost buckle is 
due to its presence. ( Courtesy of 
B. P. Kaufmann. ) 



is called a shift. If, in the heterozygote for 
a shift, the homologs pair up and a crossing 
over occurs in the region of the shift, a sec- 
tion of one of the crossovers will be in 
duplicate, as can be seen by tracing the re- 
sultant strands. 

Two breaks in one chromosome and one 
in a nonhomolog can result in the interstitial 
piece of the first chromosome being inserted 
into the second. This result is called trans- 
position. A transposition-containing chro- 
mosome can occur in subsequent generations 
not with the nonhomologous, deficient chro- 
mosome from which the piece was trans- 
posed, but with two normal chromosomes 
of that type. In this way an individual is 
produced containing a pair of normal hom- 
ologs and a part of the normal homolog 
present in hyperploid condition in a non- 
homolog. 

The preceding indicates how the same 
type of structural change — duplication — can 
result from different types of breakage 
events. For this reason, one cannot always 
specify the particular number of nonrestitut- 
ing breaks originally involved by observing 
the resultant rearrangement and, therefore, 
the explanation proposed is always the sim- 



plest one. Note also that loss of an entire 
chromosome can occur after breakage; thus, 
not all such losses come from nondisjunc- 
tion. Contrary to nondisjunction, however, 
breakage events cannot produce trisomies. 

Cytogenetic Detection of Structural Changes 

The question of how structural changes in 
chromosomes are detected may have arisen 
during the preceding discussions. Such mu- 
tants may be detected initially by cytological 
examination, or they may be noted first by 
their effects on the phenotype when genetic 
tests are made. Thus, detection and identi- 
fication of structural changes can be made 
cytologically, or genetically, or by a com- 
bination of both methods. 

When heterozygous, deficiencies can some- 
times be recognized genetically since they 
permit the expression of all genes which are 
hemizygous in the nondeficient chromosome. 
Inversions and translocations can be sus- 
pected when mutant heterozygotes show a 
marked reduction in offspring carrying cross- 
overs. Using appropriate genetic markers, 
inversion homozygotes show some genes in 
the reverse of normal order, whereas in 
heterozygotes or homozygotes for transloca- 



Struct iircd Changes in Chromosomes 



175 



tions genes normally not linked are found 
linked. Sometimes a cytological study is 
preceded by genetic studies indicating the 
class of structural change involved and the 
particular chromosome (s ) affected. Of 
course, detailed knowledge of the cytological 
appearance of the normal genome is a pre- 
requisite for such work. 

The prophase of meiosis of some organ- 
isms and the giant salivary gland chromo- 
somes of Diptera are particularly suited for 
cytological studies, because in both cases 
synapsis between homologs helps locate the 
presence, absence, or relocation of chromo- 
some parts. For example, inversion hetero- 
zygotes show either a reversed segment 
which does not pair with its nonreversed 
homologous segment (if the inversion is 
small), or (if the inversion is larger) show 
one homolog twisting in order to synapse 
( Figure 1 2-7 ) . A deficiency-heterozygote 
will buckle in the region of the deficiency. 
Since a chromosome with a duplication may 
also buckle when heterozygous, careful cyto- 
logical study is needed to distinguish this 
case from deficiency (see Figure 12-8). 
Heterozygotes for reciprocal translocations 



(Figure 12-9) show two pairs of nonho- 
mologous chromosomes associated together 
in s) napsis. 

The present discussion should suffice as 
an introduction to the origin, nature, and 
consequences of the more common types of 
structural changes in chromosomes and to 
the methods used in identifying such mu- 
tants. 



i IGURE 12-9. Heterozygous reciprocal trans- 
location in corn (pachynema) (courtesy of M. 
M. Rhoades) and Drosophila (salivary gland) 
( courtesy of B. P. Kaufmann ) . 



\ 



) 





*^t*|^* £ 




I7(> « HAP1 EB 12 

SUMMARY AND CONCLUSIONS 

Structural change in chromosomes is a type o! mutation involving the gain, loss, or 
relocation of chromosome parts. All such mutations require chromosome or chromatid 
breakage. Since proximitj favors union, most of the ends produced hy breakage 
restitute. Unions occur mainly during interphase. Noniestitutional unions produce 
structural changes ill chromosomes. The occurrence of one. two. or three nonrestituting 
hreaks in one or two chromosomes is discussed in relation to the production of whole- 
chromosome losses, deficiencies, duplications, inversions, translocations, shifts, and 
transpositions. 

Chromosomes that have undergone structural change may be euploid or aneuploid. 
The cells in which these mutations arise are euploid but can become aneuploid follow- 
ing mitosis, segregation, or crossing over. The structural changes most likely to be 
retained in the population are the smallest ones; those changes which directly or in- 
directly cause an increase in gene number are most likely to be important in evolution. 



REFERENCES 

Beam. A. G.. and German III, J. L., "Chromosomes and Disease," Scient. Amer., 
205. No. 5:66-76, 1961. 

Muller. H. J.. 'The Nature of the Genetic Effects Produced by Radiation," in Radiation 

Biology, Hollaender, A. (Ed.), New York: McGraw-Hill, 1954, Chap. 7, pp. 

351-473. 
Patau. K., "The Origin of Chromosomal Abnormalities," Pathologie-Biologie, 11:1163- 

1 170, 1963. 
Russell, L. B.. "Chromosome Aberrations in Experimental Mammals," Progress in 

Medical Genetics, 2:230-294, 1962. 



QUESTIONS FOR DISCUSSION 

12.1. The terms euploid and aneuploid (hypo- or hyperploid) have been applied both 
to individual chromosomes and to whole nuclei. Give an example of: 

(a) A hypoploid chromosome in a euploid nucleus 

(b) A hyperploid chromosome in a hyperploid nucleus 

(c) An aneuploid nucleus containing all structurally normal chromosomes 

12.2. As used on p. 170, what does the term eutelomeric mean? Name two types of 
aneutelomeric chromosomes. 

12.3. Given the chromosome AB/CDE/F.GHI/J, where the period indicates the 
centromere and the slanted lines the positions of three simultaneously produced 
breaks, draw as many different outcomes as possible. Indicate which one is 
most likely to occur. 

12.4. In Drosophila, the loss of a given chromosome results in monosomy; this situa- 
tion is approximately three to five times as frequent as its gain, resulting in 
trisomy. Explain. 

12.5. Discuss the origin of monosomies among human zygotes. 



Structural Changes in Chromosomes 111 

12.6. In human chromosomes at mitotic metaphase, discuss the detectability of the 
following: 

(a) Paracentric inversion (c) Deficiency 

(b) Pericentric inversion (d) Duplication 

(e) Half-translocation 

12.7. What advantages may inversion provide? 

12.8. What characteristics of cells undergoing oogenesis favor the production and 
viable transmission of half-translocations? 

12.9. In Drosophila, a male, dihybrid for the mutants bw and ,s7, when back-crossed 
to bw bw st st, normally produces offspring whose phenotypes are in a 1:1:1:1 
ratio. On exceptional occasions, this cross produces offspring having only two 
of the four phenotypes normally obtained. How can you explain such an 
exception? 

12.10. Is the telomere a gene? Why? 

12.11. Explain how you could cytologically determine the position of the locus for 
white on the X chromosome of Drosophila by each of the following: 

(a) deficiencies of various sizes 

(b) inversions of various sizes 

(c) various reciprocal translocations 

12.12. Suppose you had a self-maintaining strain of Drosophila in which all females 
were yellow-bodied and males, grey-bodied. How would you explain this con- 
sistency if the egg mortality were always 50%? Low, as it is normally? How 
would you test your hypothesis cytologically? 

12.13. (a) Several X-linked mutants in Drosophila cause notched wings. One of 
these mutants is lethal in the male and also in the mutant homozygote female. 
How do you suppose such a homozygote is produced? 

(b) A female heterozygous for this mutant (N/+) is mated to a fa/Y male. 
In F x all sons are normal, half the daughters are normal, and half are both 
notched and faceted. Explain this result showing how you might test your 
hypothesis. 

12.14. Make a diagram of the different eucentric reciprocal translocations between 
autosomes 2 and 3 in Drosophila which you would expect to be lethal in the 
following cases: 

(a) when either half-translocation is present 

(b) when one half-translocation but not the other is present 

(c) under no circumstances 

12.15. Does the absence of crossing over in male Drosophila facilitate the detection 
of heterozygous reciprocal translocations? Explain. 

12.16. Given a Drosophila heterozygous for a eucentric reciprocal translocation be- 
tween chromosomes 2 and 3 and assume both half-translocations are lethal 
when present separately. Discuss the nature of the linkage maps one would 
obtain from mating 

(a) genetically marked females of this type with appropriately marked non- 
translocation males 

(b) genetically marked males of this type mated to appropriately marked non- 
translocation females 

12.17. A chromosome A.BCDEEDCFG has a reverse repeat, or duplication, for CDE. 
Compare the stability of this chromosome with A.BCDECDEFG, which carries 
a tandem repeat, or duplication, for the same region. 



ITS ( II AIM IK 12 

I 2. is. In Drosophila each ol the genes for curly wings (Cy), plum eye color (/'///). 
hairless ( // ) and dichaete wings ( /> ) are lethal when homozygous. A curly, 
han less male mated to a plum, dichaete female produces 16 equally frequent 
types ol sons and daughters. One curly, plum, hairless, dichaete F, son is 
irradiated \uth \-ravs ami then crossed to a plum, dichaete female. Three F 2 
sons phenotypically like the father, collected and mated separately with wild- 
type females, produce the following males and females in the F 3 progeny. 

Phenotype Son I Son 2 Son 3 

C\ H 140 120 76 

( \ D 120 81 

Pm H 135 84 

Pm D 154 117 79 

Explain these results, using cytogenetic diagrams for all individuals mentioned. 

I2.1 l >. (a) Discuss the frequency of abortions in normal mothers who produce half- 
translocational children with Down's syndrome. 

(b) Would you sometimes expect the occurrence of children with Down's 
s\ndrome to he correlated with the father but not with his age? Explain. 

12.20. The Y chromosome is of different size in different phenotypically normal men. 
On the other hand, a woman with a small X chromosome is phenotypically 
defective. How can you explain the origin of such different Y and X chromo- 
somes and the difference in the way they affect the two sexes? 



Chapter *13 

RADIATION-INDUCED STRUCTURAL 
CHROMOSOME CHANGES 



I 



n the preceding chapter struc- 
tural changes in chromosomes 
were discussed with respect to 
types and consequences, but little was said 
about the events responsible for their pro- 
duction, namely, breakage and cross-union. 
Chromosomes break spontaneously; that is. 
they occasionally break in cells exposed to 
normal conditions. Because spontaneous 
breakage is relatively rare, agents that are 
able to produce great numbers of breaks are 
very useful in studies of chromosome break- 
age and its consequences. Our discussion 
in this chapter is restricted to one of these 
agents, radiation. 

The process of breaking a chromosome 
is a chemical reaction requiring energy. 
The biochemical effect of radiation depends 
upon the type and amount of energy left in 
tissue. Less energetic radiations (such as 
visible light) leave energy in the form of 
heat; more energetic radiations (such as 
ultraviolet light) leave energy in the form of 
heat and activation; the latter type of energy 
makes an electron move from an inner to 
an outer orbit of an atom. The more ener- 
getic the radiation, the greater the likelihood 
that the energy absorbed will lead to chem- 
ical change. For example, ultraviolet light 
produces more breaks in chromosomes than 
does visible light. Radiations of energy 
higher than ultraviolet light (X rays and 
gamma rays; alpha and beta rays; electrons, 
neutrons, protons, and other fast-moving 
179 



particles) are even more capable of causing 
breaks. Although such high-energy radia- 
tions also heat and activate, most of the 
energy left in the cells is in the form of ioni- 
zation, and this leads to most of the chromo- 
some breaks. Before discussing how ioniza- 
tion energy leads to breakage, we should 
first understand what ionization is and what 
its consequences are. 

Like visible and ultraviolet light, X and 
gamma rays are electromagnetic waves; how- 
ever, they have relatively shorter wave 
lengths and can penetrate tissue more deeply 
than visible or ultraviolet light. When a 
highly energetic wave is stopped (or a fast- 
moving particle is captured or slowed down ) , 
energy is absorbed by the atoms of the 
medium. This energy can cause an atom to 
lose an orbital electron, creating a charged 
particle, or ion, by the process of ioniza- 
tion. Such an electron, torn free of the 
atom, goes off at great speed and can, in 
turn, cause other atoms to lose orbital 
electrons — to be ionized. All atoms losing 
an electron, of course, become positively 
charged ions, and atoms that capture free 
electrons become negatively charged ions. 
Since each electron lost from one atom is 
eventually gained by another atom, ions 
occur as pairs. In this way a track of ion 
pairs, or an ion track, is produced which 
often has smaller side branches. The length 
of the main or primary ion track and its side 
branches and the density of ion pairs differ 
with the type and energy of the radiation 
involved. Fast neutrons make a relatively 
long, rather uniformly thick ion track; fast 
beta rays or electrons make a relatively long, 
uniformly thin or interrupted track of ions; 
ordinary X rays make a relatively short track 
sparse in ions at its origin becoming only 
moderately dense at its end. It is sufficient 
to say that all known ionizing radiations pro- 
duce clusters of ion pairs within microscopic 
distances. In other words, no amount or 



180 



CHAPTER 13 




kind of high-energy radiation presently 
known can produce only single ions, or 
single pairs of ions evenly spaced over 
microscopic (hence, relatively large) dis- 
tances. Since one ion or a pair sufficiently 
separated from the next does not exist, the 
genetic effects of ionization must be deter- 
mined from the activity of clusters of nega- 
tively and positively charged ions Ions 
undergo chemical reactions to neutralize 
their charge to reach a more stable con- 
figuration. It is during this process that ion 
clusters are able to produce chromosome 
and chromatid breaks (Figure 13-1). 

The amount of ionization produced by 
radiation is measured in terms of an ioniza- 
tion unit called the roentgen, or /• unit, one 
r being equal to about 1.8 X 10° ion pairs 
per cubic centimeter of air. A sufficiently 





1 


V 


.*' . 








4 - 
ft 


%' 


V 




+ 






*-*\ 






B 







* V. • 


^ 




c'- 



figure 13—1. Structural changes X-ray-induced (75-150 r) in normal human male fibro- 
blast-like cells in vitro. Arrows show: (A) broken chromosomes, (B) translocation {cen- 
ter) and dicentric (lower left), (C) ring chromosomes. A, B are in metaphase (see Fig. 
11-9); C is late prophase. (Courtesy of T. T. Puck, Proc. Nat. Acad. 5c/., U.S., 44: 
776-778, 1958.) 



Radiation-Induced Structural Chromosome Changes 



181 



penetrating radiation (such as fast elec- 
trons), producing this 1.8 X 10 !) ion pairs 
in a given cm :{ of air, can also produce this 
amount in successive cm' 5 of air because only 
a very small fraction of the incident radiation 
is absorbed at successive depths. If not very 
energetic X rays are used ("soft" X rays of 
relatively long wavelength — also called Grenz 
rays), all radiation may be absorbed near 
the surface of the medium, keeping the 
deeper regions free from ionization. The 
amount of energy left at any level depends 
not only upon the energy of the incident 
radiation, but also upon the density of the 
medium through which the radiation passes. 
Thus, in tissue, which is approximately ten 
times as dense as air, a penetrating high- 
energy radiation produces about one thou- 
sand times the number of ion pairs per cm 3 
as it does in air. Knowing this, it can be 
calculated that one r (always measured in 
air) produces about 1.5 ion pairs per cubic 
micron (/a 3 ) of tissue. Since the volume 
of the Drosophila sperm head is about 0.5 
ju 3 , one r is able to produce, on the average, 
less than one ion pair in it. Since ions occur 
in clusters, one r may place dozens of ion 
pairs in one sperm head and none in dozens 
of other sperm heads. The r unit measures 
only the absorbed energy which produces 
ions; another unit, the rad, measures the 
total amount of radiant energy absorbed by 
the medium. In the case of X rays, about 
90% of the energy left in the tissue is used 
to produce ions; the rest produces heat and 
excitation. Since ultraviolet radiation is non- 
ionizing, its dosage is measured in rads and 
not r units. 

The number of chromosome breaks pro- 
duced by X rays increases linearly with the 
radiation dose (r) (Figure 13-2). This re- 
lationship means that X rays always produce 
at least some ion clusters large enough to 
cause a break. Moreover, clusters of ions 
from different tracks of ions do not combine 
their effects to cause a break. (If there were 



co7 

or 
0Q5 - 

t- 
o 

(T2 



10 20 30 40 50 60 70 80 90 100 110 120 130 
DOSAGE IN R UNITS 

figure 13-2. The relation between X-ray dos- 
age and the frequency of breaks induced in 
grasshopper chromosomes. (Courtesy of J. G. 
Carlson, Proc. Nat. Acad. Sci., U.S., 27:46, 
1941.) 



such cooperation between clusters, the break 
frequency at low doses would be lower than 
what has been found because of the waste 
of clusters too small to break; the frequency 
at higher doses would be higher because of 
the cooperation among such clusters.) Cer- 
tain radiations, like fast neutrons, produce 
fewer breaks per r than X rays because one 
r of these radiations produces larger — and, 
hence, fewer — clusters of ions than do X 
rays. These larger clusters more often ex- 
ceed the size needed to produce a break, and 
therefore, are relatively less efficient in this 
respect. 

Ion clusters can produce breaks either di- 
rectly by attacking the chromosome itself, 
or indirectly by attacking oxygen-carrying 
molecules (which, in turn, react with the 
chromosomes) or other chemical substances 
(which, in turn, affect the chromosome or 
oxygen-carrying molecules). In any case, 
this indirect pathway must be of nearly sub- 
microscopic dimensions; otherwise, differ- 
ent ion clusters would be able to cooperate 
in causing breakage. Thus, only ion clus- 
ters in or very close to the chromosome can 
produce breaks in it, as has been visibly 
demonstrated by using beams of penetrating 



L82 



CHAPTER 13 



radiation of microscopic diameter. Such a 
beam passing through a metapliase chromo- 
some can break it. but fails to do SO when 
directed at the protoplasm adjoining the 
chromosome. 

From what has been stated, it is reason- 
able to assume that the number of breaks 
produced by a given dose of a certain radia- 
tion depends upon the volume which a chro- 
mosome occupies. This volume is different 
at different times in the nuclear cycle (for 
example, it changes during chromosome 
replication). Because of variations in poly- 
nemy or gene activity, the same chromo- 
some can occupy different volumes in dif- 
ferent tissues of an individual and the vol- 
ume of the same sex chromosome can be 
different in the two sexes. Because break- 
age requires energy, it is also reasonable to 
assume that the number of breaks indirectly 
produced increases if, during irradiation, 
either the amount of oxygen is increased, or 
the cell's reducing substances are poisoned. 
And conversely, replacement of oxygen by 
nitrogen during irradiation reduces the num- 
ber of breaks produced. 

After this preliminary discussion of some 
of the factors that influence the production 
of radiation-induced breaks, we are ready to 
consider the factors that influence the fate 
of the ends produced by breakage. Just as 
breakage involves a chemical reaction, so 
does the union between two sticky ends. 
The joining of break-produced ends appar- 
ently involves adenosine triphosphate and 
protein synthesis. 1 Joining is enhanced by 
the oxygen (and inhibited by the nitrogen) 
present after irradiation. Accordingly, resti- 
tution is prevented if nitrogen replaces oxy- 
gen after irradiation, thus increasing the 
time that ends from the same break stay 
open, and. therefore, the chance for cross- 
union when the supply of oxygen is later 
resumed. (Note that the presence of oxygen 

1 See J. G. Brewen (1963). 



has two contrary effects on rearrangement 
frequency — during irradiation it increases 
the number iA' breaks, whereas after irradia- 
tion it increases restitution. ) 

Since, under given conditions, the num- 
ber of breaks increases linearly with an ion- 
izing dose — each part of the dose independ- 
ently producing its proportional number of 
breaks — clearly, the number of breaks pro- 
duced is also independent of the rate at 
which a given total dose is administered. 
It also follows that all structural changes in 
chromosomes resulting from single breakages 
are also independent of the radiation dose 
rate. Radiations such as fast neutrons which 
produce long and dense ion tracks can fre- 
quently induce two chromosome breaks with 
the same track. In this case, if the same 
chromosome — having folded or coiled tightly 
— is broken twice by being twice in the path 
of the track, then large and small structural 
changes of inversion, deficiency, and dupli- 
cation types can be produced. The fre- 
quency of these rearrangements increases 
linearly with fast neutron dose and is inde- 
pendent of the dose rate. 

A single fast neutron-induced track of ions 
can also break two different chromosomes 
when chromosomes are closely packed to- 
gether, as they are in the sperm head. The 
linear increase with dose in the frequency 
of reciprocal translocations obtained after 
sperm are treated with fast neutrons pro- 
vides evidence for concluding — as was done 
in Chapter 1 2 — that proximity of sticky ends 
favors their union. Such a linear dose-effect 
can be obtained only if both breaks are pro- 
duced by the same track and if the broken 
ends capable of exchange union are located 
near each other — broken ends produced by 
different tracks being too far apart. 

When ordinary X rays are employed, 
however, the clusters are smaller, and the 
track of ions is shorter than fast neutron 
tracks. Accordingly, two breaks in the same 
chromosome are produced by the same 



Radiation-Induced Structural Chromosome Changes 



183 



X-ray ion track less frequently, and if they 
do occur, they are usually quite close to- 
gether. Note, however, that two breaks oc- 
curring within submicroscopic distances in 
successive gyres of a coiled chromosome 
produce structural changes whose size ranges 
only from minute to small. Nevertheless, a 
small proportion of single X-ray tracks — in 
the treatment of sperm, for example — do 
cause two breaks, each in a different chro- 
mosome. Therefore, for X-ray doses that 
produce fewer than two tracks per sperm, 
gross chromosomal rearrangement frequency 
increases linearly with dose. So, there is 
actually no dose of X rays which does not 
have some chance of producing a gross re- 
arrangement. In other words, no matter 
how small a dose of ionizing radiation is 
received, the possibility of a chromosomal 
break and a gross chromosomal mutation 
always exists. 

In the case of X rays or fast electrons, 
two breaks that occur in the same nucleus 
usually result from the action of two ion 
clusters, each derived from a different, inde- 
pendently arising track, so that each break 
is induced independently. Fast electron or 
X-ray-induced, two-break gross rearrange- 
ments of this origin are dose dependent, for 
when a small enough dose is given, a nucleus 
is traversed by only one track, and only 
one-track — not two-track — gross rearrange- 
ments can result. But when the dose is large 
enough for a nucleus to be traversed by two 
separate tracks, the two breakages required 
for two-break gross rearrangements can be 
produced independently. Therefore, the 
higher the dose of X rays used, the greater 
the efficiency in producing multi-break gross 
rearrangements caused by breaks independ- 
ently induced by separate tracks. Accord- 
ingly, for doses causing some cells to experi- 
ence two such independently produced 
breaks and higher doses, the frequency of 
these mutations increases more than in di- 
rect proportion to the amount of dose. One 



example is the exponential rise in the fre- 
quency of reciprocal translocations obtained 
after treating sperm in inseminated Dro- 
sophila females with increasing dosages of 
fast electrons (Figure 13-3, curve T). 

X-ray-induced rearrangements involving 
two (or more) breaks induced by separate 
tracks also depend upon the rate at which 
a given dose is administered. When a suit- 
ably large dose is given over a short interval, 




i i i i — i — i — r 

10 14 18 22 26 30 34 38 
DOSE IN RADS(XIOO) 

figure 13-3. Percentage of mutations, ±2X 
standard error, recovered from Drosophila 
sperm exposed to different dosages of 18 mev 
electrons. The sex-linked recessive lethal fre- 
quencies (L) are joined by solid lines and are 
adjusted for the control rale; sex chromosome 
loss frequencies (5) are connected by broken 
lines and are corrected for the control rate; 
reciprocal translocation frequencies (T) be- 
tween chromosomes II and III are connected 
by dot-dash lines. (From I. H. Herskowitz, 
H. J. Midler, J. S. Laugh lin. Genetics, 44:326, 
1959.) 



184 



( HAPTER 13 



the ends produced h\ separate breaks exist 
simultaneously and are able to cross-unite. 
But when the same dose is given more 
slowly, the pieces of the fust break ma\ 
restitute before those o\ the second arc pro- 
duced, thus eliminating the opportunity for 
cross-union. In this event, the same dose 
produces fewer gross rearrangements when 
given in a protracted manner than when 
given in a concentrated manner. Although 
this dose-rate dependence for X rays is true 
for most cells — at least during part of the 
interphase stage — it does not apply to ma- 
ture sperm of animals, probably including 
man. In these gametes and during most of 
nuclear division in other cells, the broken 
pieces cannot join each other (sec p. 166) 
and. therefore, accumulate. For this rea- 
son, it makes no difference how quickly 
or slowly the dose is given to the chromo- 
somes in such a sperm head, since the breaks 
remain unjoined at least until the sperm 
head swells after fertilization. 

As already mentioned, the spatial arrange- 
ment of chromosomes with respect to each 
other influences the number of breaks and 
the kinds of structural changes they produce. 
It should be noted that the possibilities for 
multiple breakages and for joinings are quite 
different for chromosomes packed into the 
tiny head of a sperm than they are for chro- 
mosomes located in a large nucleus. But 
even within a given type of cell, a number 
of other factors can influence breakage or 
rejoining, such as the presence or absence 
of a nuclear membrane, the degree of spiral- 
ization of the chromosomes, the stress or 
tension under which the parts of a chromo- 
some are held, the degree of hydration, the 
amount of matrix in which the genes are 
embedded, protoplasmic viscosity and the 
amount of fluid and particulate movement 
around the chromosomes, gravity, centrif- 
ugal force, and vibration. 

In cells whose chromosomes have just 
replicated and in somatic or meiotic cells 



where homologS arc synapscd, a special re- 
striction on the movements of the pieces is 
produced when only some of the apposed 
strands are broken (see p. 166). In this 
situation, the forces which keep parts of one 
strand adjacent to the corresponding parts 
of its sister or homolog may prevent the 
broken pieces from moving apart freely, so 
that the unbroken strand or strands serve as 
a splint for the broken one(s) and reduce 
the opportunities for cross-union. Many 
factors exist, therefore, which determine to 
what degree chromosome and chromatid 
fragments can move or spring apart; those 
affecting the distances between different 
chromosomes or the parts within a chromo- 
some also affect chromosome and chromatid 
breakability. 

The frequencies and types of structural 
changes depend also upon the total amount 
of chromosomal material present in the nu- 
cleus and the number and size of the chro- 
mosomes into which this material is divided. 
The rearrangements that occur in different 
cells of a single individual depend upon 
whether the cell is haploid, diploid, or poly- 
ploid, and whether or not the chromosomes 
are polynemic, are in the process of replica- 
tion, or are otherwise metabolically active. 

Radiation can produce important non- 
mutating effects upon the chromosomes by 
damaging nonchromosomal cellular compo- 
nents which, in turn, affect chromosomal be- 
havior and function. If the cells are capable 
of repairing such nonchromosomal, struc- 
tural or functional damage, they will have a 
longer time in which to repair when a radia- 
tion dose is given slowly than when given 
quickly. The most obvious example is the 
effect of radiation upon mitosis (and prob- 
ably meiosis). Cells at about midprophase 
or a later stage in nuclear division usually 
complete the process even though irradiated. 
Cells no farther advanced than about mid- 
prophase often return to interphase when 
irradiated. For this reason, ionizing radia- 



Radiation-Induced Structural Chromosome Changes 



185 



tion causes a greater degree of synchromy 
in division than occurs in the absence of the 
radiation. Accordingly, starting with a pop- 
ulation of cells in various stages of nuclear 
division, the chromosomal targets for muta- 
tion are different in the later stages of re- 
ceiving a protracted dose and of receiving a 
concentrated dose. 

The capacity to produce recoverable struc- 
tural changes is not the same in euchro- 
matic and heterochromatic chromosomal 
regions. Recovered radiation-induced struc- 
tural changes involve heterochromatic regions 
more frequently than they do euchromatic 
ones. It has not been determined whether 
this excess is due to heterochromatin hav- 
ing a greater breakability, a lesser resti- 
tutability (which might be associated with 
the general ability of different heterochro- 
matic regions to synapse with each other), 
or both. Nevertheless, in many rearrange- 
ments, at least one of the points of breakage 
is located in the heterochromatin nearest the 
centromere. This is one reason why whole- 
arm reciprocal translocation is the type most 
frequently observed. 

The present discussion has been motivated 
by the ability of energetic radiations to in- 
duce many breaks and, subsequently, many 
structural changes. The great supply of re- 
arrangements readily available via radiation 
treatment has made it possible to discover 
many of the factors influencing breakage and 
joining. Many other important discoveries 
were made possible by the study of structural 
changes, including 

1 . The genetic basis of the centromere 

2. The reduced incidence of crossing over 
near the centromere 

3. The genetic basis of the telomere 

4. The existence in some species of ge- 
netic elements (collochores) near the 
centromere of special importance to 
synapsis. - 

^ See K. W. Cooper (1964). 



X CHROMOSOME 




cv ct 
y sc br pn wrb 



Centromere 



H CHROMOSOME 



/ I0\20 30 40^50 



17? ^^ \ \ 

,60-^70^80 90 100 107 



al dp 



7 v" 



b Bl/\cn vg c 
It tk 



P* sp 



Centromere 



IE CHROMOSOME 



Q__jg_Aig__3Q_l40?t ,5Q^60 70 80 90 IQO 



D th cu sr e 



Centromere 



figure 1 3-4. Comparison of chromosome 
(hollow bar) and crossover (solid bar) maps 
in D. melanogaster. 



Perhaps the most fundamental contribution 
was the finding, via structural changes, that 
the genes have the same linear order in the 
chromosome, that is, in chromosome maps, 
as they have in crossover maps. The spac- 
ing of these, however, is different in the two 
cases (Figure 13-4). Thus, for example, 
because of the reduction in crossing over 
near the centromere, the genes nearest the 
centromere — spaced far apart in the meta- 
phase chromosome map — are found to be 
close together in the crossover map. 

Although our subject matter has been re- 
stricted to the factors influencing the origin 
and joining of breaks produced by ionizing 
radiation, these factors are expected to oper- 
ate on breaks produced by any other spon- 
taneously occurring or induced mechanism. 
For. in general, no matter how broken chro- 
mosomes are produced, all possess the same 
properties. 



186 CHAPTER 13 

SUMMARY AND CONCLUSIONS 

The components ol structural chromosome change, breakage and cross-union, are 
readilj studied through the use ol ionizing radiations. These radiations induce break- 
aye in chromosome strands primarily by the clusters of ion pairs they produce. These 

clusters form tracks ol ions whose thickness and length determine the number and loca- 
tion ol the breaks, hacks ol ions must occur very close to. or within, the chromosome 
lh.it the) break. 

The number of breaks increases linearly with radiation dose. Whether they result 
from one or from two breaks, all chromosomal rearrangements induced by a single 
ionizing track increase linearly with radiation dose, have no threshold dose, and show 
no effect from protracting or concentrating the dose. Accordingly, there is no dose 
of ionizing radiation which does not produce breaks and at least single-track-induced 
rearrangements. 

Two-or-more-break structural changes produced by ion clusters in separate, inde- 
pendently-occurring tracks increase in frequency faster than the amount of dose and 
do have a threshold dose. If joining of chromosome ends produced by breakage can 
take place during the course of irradiation, such rearrangements are reduced in fre- 
quency by protracting the delivery of the total dose. 

Since both the breakage and joining processes involve chemical changes, their fre- 
quencies can be modified by the metabolic state of the cell. All types of rearrange- 
ments are expected to be affected by: the physical and chemical state of the chromo- 
some and the amount and distribution of its euchromatin and heterochromatin; by the 
number and arrangement of the chromosomes present; by the presence or absence of a 
nuclear membrane; and by the movements of broken ends as influenced by cellular 
particles, fluids, and extracellular factors. 

REFERENCES 

Bacq, Z. M., and Alexander. P., Fundamentals of Radiobiology , 2nd Ed., New York: 

Pergamon Press, 1961. 
Bender, M. A., and Gooch, P. C, "Types and Rates of X-ray-Induced Chromosome 

Aberrations in Human Blood Irradiated in Vitro," Proc. Nat. Acad. Sci., U.S., 

48:522-532, 1962. 
Brewen, J. G.. "Dependence of Frequency of X-ray-Induced Chromosome Aberra- 
tions on Dose Rate in the Chinese Hamster," Proc. Nat. Acad. Sci., U.S., 50:322- 

329, 1963. 
Chu, E. H. Y., Giles, N. H., and Passano, K., "Types and Frequencies of Human 

Chromosome Aberrations Induced by X-rays," Proc. Nat. Acad. Sci., U.S.. 47: 

830-839. 1961. 
Cooper, K. W., "Meiotic Conjunctive Elements Not Involving Chiasmata.' Proc. Nat. 

Acad. Sci.. U.S.. 52:1248-1255. 1964. 
"Ionizing Radiation," Scicnt. Amer., 201 :No. 3 (Sept.), 1959. 
Muller. H. J.. "General Survey of Mutational Effects of Radiation," in Radiation 

Biolo^x and Medicine, Claus, W. D. (Ed.), Reading, Mass.: Addison-Wesley, 

Chap. 6, pp. 145-177. 1958. 
Puck, T. T., "Radiation and the Human Cell," Scient. Amer., 202, No. 4:142-153, 

1960. 
Sobels, F. H.. Repair from Genetic Radiation, New York: Pergamon Press, 1963. 
Sparrow, A. H.. Binnington. J. P., and Pond, V., Bibliography on the Effects of Ionizing 

Radiations on Plants, 1896-1955, Brookhaven Nat. Lab. Publ. 504 (L-103), 1958. 



Radiation-Induced Structural Chromosome Changes 



187 



Lewis John Stadler (1896-1954) 
is noted for his studies on the nature 
of mutation and of the gene (see p. 
ix). He and H. J. Muller discov- 
ered independently the mutagenic 
effect of X rays. {From Genetics, 
vol. 41, p. 1, 1956.) 




QUESTIONS FOR DISCUSSION 

13.1. After both are exposed to the same amount of radiation why should tissue, 
which is only about ten times as dense as air, contain about one thousand times 
more ions than air? 

13.2. What evidence can you give to support the view that the ions causing breakage 
need not always attack the chromosome directly? 

13.3. Does the observation that the volume of a chromosome is variable under dif- 
ferent conditions mean that it has an inconstant gene content? Explain. 

13.4. Do you suppose that chromosomes exposed to X rays are more likely to under- 
go structural change when they are densely spiralized than when relatively un- 
coiled? Why? 

13.5. Discuss the role of heterochromatin in changes involving chromosome number 
and chromosome shape. 

13.6. Do you suppose that the oxygen content of a space capsule can affect the 
mutability of Drosophila passengers? Explain. 

13.7. Discuss the relative efficiency, per r, of small doses of X rays and of fast neu- 
trons in producing structural changes in chromosomes. 

13.8. Do you suppose that the mutability of ultraviolet light threatens man's survival? 
Explain. 



1SS CHAPTER 13 

] 3.9. Compare the number and fate of breakages induced b) the same dose of X rays 
administered to: 

(a) A polyploid and a diploid liver cell m man 
(h) A diploid neuron in man and Drosophila 

(e) A sperm and a spermatogonium in man 

13.10. Discuss the importance of the nonmutant effects ol a given dose of radiation 
upon the mutation frequency induced b\ a subsequent radiation dose. 

13.11. Using Figure 13-4. discuss the likelihood of crossing over in different regions 
of the X chromosome of D. melanogaster, 

13.12. Compare the roentgen unit with the rad unit. 

13.13. What specific aspects of our present environment tend to reduce the number 
of mutations induced by penetrating radiations from the number induced when 
man first evolved? 



Chapter *14 

POINT MUTATIONS 



W: 



e have already found that 
the mutational unit of the 
genotype may be a whole 
genome, a single chromosome, or a part of 
a chromosome. Perhaps a study of these 
units will reveal more about the mutational 
characteristics of a single gene; perhaps 
the recombinational properties of individual 
genes will illuminate this area of investiga- 
tion. Let us consider what we already know 
about the mutation of single genes — the class 
of mutation that is probably the most impor- 
tant in evolution because it causes the small- 
est shift in gene balance. 

All chromosomes are linear and un- 
branched whether or not they have under- 
gone segmental rearrangement by crossing 
over or breakage. This linear arrangement 
could be due to the linkage of gene to gene 
directly, or indirectly by a nongenetic ma- 
terial which connects adjacent genes. In 
either case, the fact remains that a chromo- 
some is invariably either a rod or a ring, 
providing almost conclusive evidence that a 
gene cannot be joined to other genes at more 
than two places, and that a mutation which 
permits a gene to be joined to more than two 
others cannot occur spontaneously or be in- 
duced. That this type of mutation is never 
observed regardless of the organism studied 
can be interpreted to mean that genes never 
had this property or that all existing genes 
have lost this property. We are led to con- 
clude, therefore, that all interstitial genes are 
bipolar, and that mutation is incapable of 
189 



causing the gene to be more than bipolar. 
After chromosome breakage, the "stick- 
iness" of the new ends is evidence that al- 
most all mutations retain the bipolarity of 
genes. In some relatively rare cases, how- 
ever, break-produced ends (broken ends) are 
known to become permanently unsticky or 
healed, so that mutation from bipolarity to 
unipolarity does occur. That mutation can 
change genes from a bipolar to a unipolar 
type, or vice versa, is evidenced also by the 
presence of telomeres — unipolar genes which 
seal off the normal ends of chromosomes. 

The chromosomal change from bipolarity 
to unipolarity occurs regularly in the life 
history of certain animals. In particular 
species of the roundworm Ascaris, for ex- 
ample, nuclei which remain in the germ line 
have a single pair of chromosomes. When 
the nuclei first enter the somatic line, how- 
ever, these chromosomes break up into a 
number of small linear fragments whose ends 
are sealed off and behavior during mitosis 
is normal — normal mitotic behavior being 
possible because a germ line chromosome 
has numerous centromeres along its length 
(each surviving fragment of the chromosome 
in a somatic cell has at least one). In the 
germ-line polycentric chromosome all centro- 
meres but one are suppressed. Because 
chromosome fragmentation in Ascaris takes 
place only in somatic cells, these polarity 
changes can be attributed to some physiolog- 
ical difference between cells entering the so- 
matic line and cells remaining in the germ 
line. These polarity changes should be con- 
sidered recombinational rather than muta- 
tional events because the changes from 
bipolarity to unipolarity are numerous, si- 
multaneous, and normal — therefore lacking 
the novelty of mutations. 

Although mutations which change polarity 
from bipolarity to unipolarity apparently oc- 
cur, no unambiguous case has ever been 
reported of the reverse, that is, of a muta- 



190 



CHAPTER 14 



tion from unipolarity to bipolarity. Since 
the chance of detecting and proving a change 
from uni- to bipolarity is verj small indeed, 
the occurrence of such a change cannot, at 
present, be denied with certainty. Do muta- 
tions to nonpolaritv occur'.' It is evident that 
a unipolar or bipolar gene that mutates to a 
Donpolar alternative must necessarily drop 
out of the chromosomal line-up. If this hap- 
pens, the freed, not-at-all-sticky gene will 
not he linked to any chromosome. Since 
no evidence has yet been presented for the 
existence of genetic material liberated from 
its chromosomal locus in this way, we cannot 
give an affirmative answer at this point. 

The gene was first identified in sexually 
reproducing individuals whose chromosomes 
synapse during meiosis. Synapsis results 
from the attraction between different seg- 
ments of one or more chromosomes. That 
different degrees of specific attraction exist 
between genes is illustrated by the fact that 
genes located in heterochromatin synapse 
much less specifically than those found in 
euchromatin. Specific genes (such as one 
in maize called asynaptic) are known which 
not only lack synaptic attraction for their 
alleles but also destroy this attraction be- 
tween pairs of genes at other loci, or cause 
general desynapsis. The occurrence of col- 
lochores — genes which assist in pairing — has 
already been mentioned in Chapter 13. 
Corresponding euchromatic loci located in 
homologous chromosomes synapse with each 
other whether or not the particular alleles 
contained are identical or different. Yet 
euchromatic genes in nonhomologs do not 
usually synapse with each other, although it 
is presumed that some presently nonallelic 
genes were previously allelic. Consequently, 
mutation must be capable of changing the 
synaptic specificity of a gene; and it must 
follow, at least in a general way, that iden- 
tical genes attract each other more than non- 
identical ones. 

Since at least some genes have multiple 



alleles, it is clear that different forms of a 
gene do exist, and mutations of such genes 
are not explicable merely in terms of their 
complete loss or inactivation. Since some 
mutations produce no visible change in the 
handing pattern of salivary gland chromo- 
somes of Drosophila, mutations involving 
but a single gene, that is, gene mutations, 
can be submicroscopic. At present, we can 
only detect gene mutations by the pheno- 
typic changes they produce. Consequently, 
the characteristics of gene mutation must be 
determined from the phenotypic changes 
produced by recombinationally detected 
genes. Accordingly, we are unable to de- 
termine from such phenotypic changes 
whether gene mutation involves the recom- 
binational gene in toto, a one portion or site 
within it. or many different sites within it. 
If gene mutation involves a change in the 
entire gene, then the material composition 
of the genes detected by recombination and 
by mutation would be identical. If, on the 
other hand, the recombinationally detected 
gene contains one or more sites at which 
mutation can occur, the basic recombina- 
tional unit of genetic material would be larger 
than the basic mutational unit. Until such 
time as critical evidence is obtained to the 
contrary, we have no choice other than con- 
tinuing to accept the mutational and recom- 
binational genes as materially equivalent, an 
assumption (Chapter 3, p. 36) which is in 
accord with the law of parsimony. 

As mentioned in the first chapter, any 
given gene is rather stable, having been faith- 
fully replicated many thousands of times be- 
fore a detectable mutation occurs. The 
greater the sensitivity of our tests for de- 
tecting mutations, however, the larger is the 
frequency of mutation observed (recall the 
detection of isoallelism, p. 59). It is rea- 
sonable to assume therefore that transmis- 
sible modifications of single genes do occur 
which escape our present means of detection. 
Nevertheless, within the limits of our present 



Point Mutations 



191 



methods of analysis, the gene appears to be 
a very stable entity. 

Consider the following method for obtain- 
ing information with regard to gene muta- 
tion. All mutants involving the one or more 
genes being investigated are collected and 
then analyzed. Some mutants involving a 
given locus prove to be based upon changes 
in the number of whole chromosomes; others 
prove to be associated with gross or small 
chromosomal rearrangements. All these 
mutants are eliminated from further consid- 
eration even though gene mutation may also 
have occurred. Insofar as feasible, all ge- 
netic and cytological tests known are applied 
to eliminate mutants involving the minutest 
chromosomal rearrangements including, for 
example, tiny duplications or deficiencies. 
All, or a considerable number, of the mu- 
tants remaining can then be assumed — for 
lack of evidence to the contrary — to have 
resulted from mutations involving either a 
single gene (gene mutations) or at most 
only a few genes (intergenic mutations). 
Each of the remaining mutants behaves as 
though it resulted from a change at a single 
point in the crossover and cytological maps 
and is, therefore, called a point mutant. 
Since at this point we have no criterion for 
differentiating between a mutant involving 
only one gene (including its complete loss) 
and one involving a few genes, the entire 
heterogeneous class of point mutants will 
have to be studied in the hope of revealing 
some of the characteristics of gene mutation. 
Consider some of the characteristics of 
spontaneous and induced point mutations. 
Since point mutation of a vast number of 
different genes occurs, this process is not 
restricted to a very limited type of gene. 
Although the conditions causing point muta- 
tion might be of such a nature that, in the 
diploid cell, both members of a pair of alleles 
tend to mutate at the same time, actually the 
evidence is that only one gene of the pair 
is affected. Because only one member of 



a pair of genes in a nucleus mutates, point 
mutation must be a very localized, submicro- 
scopic event. 

If point mutation usually involved either 
a series of stable gene changes or an in- 
stability of the gene extending over more 
than one cell generation, the resultant mu- 
tants would usually occur in clusters and 
within a cluster the same gene might not 
always mutate to the same allele. But many 
point mutants occur singly. Moreover, 
those which do appear in a cluster often 
seem to be identical. Such a cluster can 
usually be explained by assuming a single 
cell has undergone mutation, having divided 
a number of times before the tests to detect 
the mutants were performed. Although 
such data do not prove that point mutation 
is instantaneous, they indicate that it is usu- 
ally completed within one cell generation 
and the change in this respect is quick more 
than it is gradual. The number of point 
mutations obtained from X-ray or ultra- 
violet ray treatments is reduced, however, 
if posttreatment with certain types of visible 
light or chemical substances is given im- 
mediately (but not if such treatment is post- 
poned for some hours). Such immediate 
posttreatments produce photo- or chemo- 
recovery from point mutation and prove that 
the point mutation process often does not 
occur or is not completed for some minutes. 
Certain chemical changes, which themselves 
may or may not be mutational, can lead to 
other, genetic changes such as breakage. If 
the first changes are repaired before they can 
induce the second, an apparent recovery 
from mutation is observed. Only after the 
point mutation process is completed is the 
new genetic alternative just about as stable 
as the old. 

Because point mutants are just about as 
stable as their parent genes or other genes 
in the genotype, it does not necessarily mean 
that all allelic and nonallelic genes have 
the same spontaneous mutation frequency. 



192 



CHAPTER 14 



Study o\ a representative sample ol specific 
loci in Drosophila reveals an average ol one 

point mutation at a given locus in each 200,- 
000 germ cells tested. In mice the per locus 
frequency is about twice this, or one in 
100,000. In man. by scoring the mutants 
detected in heterozygous condition, the per 
locus rate is found to be one per 50.000 to 
100,000 germ cells per generation. Within 
a species, different loci have about the same 
order of mutability. Even though some 
genes are definitely more mutable than 
others, the average spontaneous point muta- 
tion rate per genome per generation can be 
estimated for Drosophila, mouse, and man. 
In one Drosophila generation, one gamete 
in twenty (or one zygote in ten) contains 
a new detectable point mutant. In mice, this 
frequency is about one in ten gametes, 
whereas in man it is about one in five gam- 
etes (or two in five zygotes). 

The point mutations which occur spon- 
taneously — that is, under natural conditions 
— bear no obvious relation to the environ- 
ment, either with respect to the locus af- 
fected or the type of alternative produced. 
Modifications in the environment do, how- 
ever, influence point mutation frequency. 
For example, in the range of temperatures 
to which individuals are usually exposed, 
each rise of 10 C produces about a fivefold 
increase in point mutation frequency. The 
magnitude of this increase is similar to, al- 
though somewhat greater than, that obtained 
with an increase in temperature in ordinary 
chemical reactions. Violent temperature 
changes in either direction produce an even 
greater effect upon point mutation fre- 
quency. Actually, detrimental environmen- 
tal conditions of almost any kind increase 
point mutation frequency. 

Physical and chemical agents which raise 
the mutation frequency enormously are 
called mutagens. All high-energy radia- 
tions (see Chapter 13) are mutagenic (see 
Figure 13-3) as are many highly reactive 



chemical substances including: mustard gas 
and its derivatives; peroxides; epoxides; and 
carbamates. The point mutation frequencies 
obtained with radiation and certain chemical 
mutagens can be 150 times the spontaneous 
frequency. One speaks of a "spectrum of 
spontaneous point mutations in that, as men- 
tioned, certain loci are normally somewhat 
more mutable than others. The loci affected 
and the types of mutant alternatives pro- 
duced by ionizing radiation are not radically 
different from those involved in spontaneous 
mutation. That these radiations produce a 
mutational spectrum much like the sponta- 
neous one is expected, since radiant energy 
is more or less randomly distributed in the 
nucleus and generally enhances many differ- 
ent kinds of chemical reactions. The point- 
mutational spectra for different chemicals 
are somewhat different from each other as 
well as from the spectra induced by radia- 
tion mutagens or by spontaneous factors. 
These differences can be attributed to the 
nonrandom penetration of these chemical 
substances into the nucleus, or to their spe- 
cific capacities for combining with different 
nuclear chemicals, or both. Nevertheless, 
the frequency of point mutation, which in- 
creases linearly with the dose of ionizing 
radiation (although the frequency is in- 
fluenced by the amount of oxygen present), 
probably also increases linearly with the nu- 
clear dose of many different chemical muta- 
gens. So point mutation probably has no 
threshold dose with chemical mutagens, and 
the number of point mutations produced by 
a given total dose is constant, other things 
being equal, regardless of the rate of de- 
livery. 

For ultraviolet light — which is not a highly 
energetic radiation — the situation is differ- 
ent. Here the probability for the individual 
unit or quantum of energy inducing point 
mutation is considerably less than 100 per 
cent. Moreover, because several quanta can 
cooperate to produce mutation, ultraviolet 



Point Mutations 



193 



induced point-mutation frequency increases 
taster than linearly with dose — at least for 
low doses — and an attenuated dose is less 
mutagenic than a concentrated one. 

Point mutation is not restricted to the 
genes of any particular kind of cell, occur- 
ring in males and females, in somatic tissues 
of all kinds, and in the diploid and haploid 
cells of the germ line. Later stages in gam- 
etogenesis and very early developmental 
stages — perijertilization stages — are found 
to be relatively rich in spontaneous point 
mutations. Despite very great differences in 
life span, one does not find correspondingly 
great differences in the spontaneous germ 
line mutation frequencies of flies, mice, and 
men. This similarity in mutation frequency 
is not surprising if most of these mutations 
occur in the perifertilization stages, since 
each of these organisms spends a comparable 
length of time in these stages. Still another 
similarity among these species is the com- 
parable number of cell divisions required for 
each to progress from a gamete of one gen- 
eration to a gamete of the next. In fact, the 
differences in mutation frequency for these 
organisms are approximately proportional 
to the differences in the number of germ 
cell divisions per generation. 

When during the history of the gene does 
mutation occur? The finding that the point 
mutation frequencies in Drosophila, mouse, 
and man are proportional to the number of 
cell divisions they undergo suggests that some 
of these mutations occur at synthesis of the 
new gene, although the experimental results 
do not specify whether it is the old or the 
new gene that mutates. Aging of spermatids 
and sperm of Drosophila is known to in- 
crease the point mutation frequency. Since 
the viability of these cells is not impaired 
when aneuploid, the increase in point muta- 
tions may be due to an effect upon the old, 
physiologically quiescent gene, implying that 
point mutational changes can occur while a 
gene is linearly attached to its neighbors. 



The larger number of mutations obtained 
from aged cells may also be explained as 
resulting from a mutagen accumulated over 
a period of time which acts on the old or 
the new gene once gene replication is re- 
sumed. The possibility also remains that 
changes can occur in the steps leading to 
gene synthesis — before the new gene is com- 
pleted and attached to its linear neighbors; 
such changes could be scored later as point 
mutants. 

Phenotypic Effects of Point Mutants 

The biological fitness of a mutant gene — 
pure or hybrid — is best described in terms 
of its effect upon the organism's ability to 
produce surviving offspring, that is, upon re- 
productive potential. This potential includes 
the mutant-carrying individual's capacity to 
reach the reproductive stage and its fertility 
and fecundity during this period, as well as 
the viability of its offspring until sexual ma- 
turity. Although each mutant has many 
phenotypic effects, point mutants with small 
phenotypic effects occur much more fre- 
quently than those with large effects. For 
instance, pure (homo- or hemizygous) mu- 
tants which lower the viability of males with- 
out being lethal are at least three to five 
times more frequent than those which are 
recessively lethal (Figure 13-3). 

The vast majority of point mutants have 
a detrimental effect on the reproductive 
potential; beneficial point mutants are 
extremely rare. In terms of the past 
evolutionary history of a species, it is under- 
standable that in the great majority of cases, 
mutants affecting a trait or organ cause its 
degeneration. All the genotypes in a species 
have been subjected to selection for many 
generations, those producing the greatest re- 
productive potential having been retained. 
Although point mutation at any locus is a 
rare event, many of the possible alternatives 
for each gene must have occurred at least 
several times in past history. Of these 



194 



CHAPTER 14 



alternatives, only the more advantageous 

alleles were retained, and these are the ones 
found in present populations. So. a point 
mutation today is likely to produce one of 
the genetic alternatives which occurred also 
in the past but had been eliminated beeause 
ol its lower biological fitness, that is, its 
lower reproductive potential. It should be 
realized, moreover, that reproductive poten- 
tial is the result of coordinated action of the 
whole genotype. The genotype may be 
likened to the machinery that makes modern 
automobiles — the automobile representing 
the phenotype — with the environment fur- 
nishing the necessary raw materials. Pres- 
ent genotypes, like the machines that manu- 
facture automobiles, are complex and have 
had a long evolutionary development. The 
chance that a newly-occurring point muta- 
tion will increase reproductive potential is 
just as small as the chance that a random 
local change in the present machinery will 
result in a better automobile. 

The differences between the phenotypic 
effect of a point mutant and its normal alter- 
native can be studied by adding more repre- 
sentatives of the mutant allele to the geno- 
type and examining the effect. In Drosoph- 
ila, for example, the normal fly has long 
bristles when the normal, dominant gene 
bb + is present. A mutant strain has shorter, 
thinner bristles because of the recessive allele 
bb (bobbed bristles), which — it should be 
recalled — has a locus both in the X and the 
Y chromosomes. We might suppose that 
the male, or female, homozygous for bb has 
bobbed bristles because this allele results in 
thinning and shortening the normal bristle. 
Since otherwise-diploid XYY males and 
XXY females can be obtained which carry 
three bb alleles, according to this view, one 
would expect the bristles formed to be even 
thinner and shorter than they are in ordinarv 
mutant homozygotes. But, on the contrary, 1 




3M 1 • 
DOSAGE OF GENES 



figure 14-1. The relationship between dos- 
age of normal and mutant genes and their 
phenotypic effect. 

in the presence of three representatives of 
bb — that is, three doses of bb — the bristles 
are almost normal in size and shape. This 
finding demonstrates that bb functions in 
the same way as bb+ does, but to a lesser 
degree. Mutants whose effect is similar but 
less than the normal gene's effect are called 
hypomorphs. Many point mutants are hy- 
pomorphs, since, in the absence of the nor- 
mal gene, additional doses cause the pheno- 
type to become more normal. 

Of the remaining point mutants, most are 
amorphs; these produce no phenotypic effect 
even when present in extra dose. One ex- 
ample is the gene for white eye (w) in Dro- 
sophila. 

Some mutants, neomorphs, produce a new 
effect — adding more doses of a neomorphic 
mutant causes more departure from normal, 
whereas adding more doses of the normal 
alternative has no effect. 

The relationship between the normal, 
wild-type gene and its hypomorphic mutants 
is indicated diagrammatically in Figure 
14-1. 1 ' The vertical axis represents pheno- 
typic effect; the normal, wild-type effect is 



1 As shown by C. Stern. 



-Adapted from H. J. Muller. 



Point Mutations 



195 



indicated by -f . The horizontal axis refers 
to the dosage of either the normal gene or 
a hypomorphic mutant. Notice that a sin- 
gle + gene itself produces almost the full 
normal phenotypic effect (and often the dif- 
ference between its effect and the normal 
effect is not readily detected) . Two -f genes 
reach the wild-type phenotypic level. In the 
case of the hypomorphic mutant, however, 
even three doses may not reach the pheno- 
typic level produced by one -(- gene (recall 
the discussion oibb). Note also that genetic 
modifiers or environmental factors, which 
can shift the position of the genes on the 
horizontal axis and thereby shift the pheno- 
typic effect, have a decreasing influence as 
one proceeds from individuals carrying only 
one dose of mutant toward individuals carry- 
ing two + genes. Natural selection would 
clearly favor alleles that result in phenotypic 
effects close to wild-type — that is, near the 
curve's plateau — for such alleles assure 
phenotypic stability. Any mutant which 
produced such a phenotypic effect would, 
in the course of time, become the normal 
gene in the population and would automat- 
ically be dominant when heterozygous with 
a hypomorphic gene alternative. This model 
illustrates how the heterozygote with one + 
and one mutant gene has practically the same 
effect as the normal homozygote, and it 
seems to best explain most cases of com- 
plete or almost complete dominance. Since 
the normal gene alternative already pro- 
duces a near-optimum phenotypic effect, this 
scheme also illustrates why, other things be- 
ing equal, so few mutants are beneficial. 

Although it is understandable from the 
preceding discussion that hypomorphic and 
amorphic mutants are usually detrimental 
when pure, one may still wonder what effects 
these mutants have when heterozygous with 
the normal gene. If the mutant is an 
amorph, the mutant heterozygote can fall 
short of producing the wild-type phenotypic 
effect and, therefore, such mutants are ex- 



pected to be sJightly detrimental when het- 
erozygous. Hypomorphs are expected to be 
less or not at all detrimental when hetero- 
zygous, at least with respect to the trait for 
which they are classified as hypomorphic. 
But since each gene affects many different 
biochemical processes, a mutant hypo- 
morphic in respect to one trait may be 
amorphic in respect to another. In Dro- 
sophila, for example, the normal allele apr+ 
which results in dull-red eye color also pig- 
ments the Malpighian tubules. One of its 
alleles, apr, causes a lighter eye color (being 
hypomorphic in this respect) but no color 
in the Malpighian tubules (being amorphic 
in this respect). 

Experience confirms the expectation that 
most "recessive" lethal point mutants — these 
are lethal when homozygous — also have 
some detrimental effect on reproductive po- 
tential when heterozygous. Such mutants 
are not completely recessive, therefore, and 
when heterozygous in Drosophila cause 
death before adulthood in about two per 
cent of individuals. Usually mutants which 
are detrimental but not lethal when pure also 
show a detrimental effect when heterozy- 
gous; this effect is somewhat less than that 
produced by heterozygous recessive lethal 
point mutants. The principles of phenotypic 
action discussed here are expected to apply 
both to spontaneous and to induced point 
mutants. 

Detection of Point Mutants in Drosophila 

We have already mentioned the existence of 
genetic methods for collecting point mutants. 
Let us now consider in some detail one ele- 
gant procedure ;; employed in Drosophila 
melanogaster for this and other purposes. 

The commonly-used technique for detect- 
ing recessive lethals is called "Base" (see 
Figure 14-2) and was designed 4 to discover 
such mutants arising in the male germ line, 

:i Invented by H. J. Muller. 

4 To replace the old "C1B" method. 



L96 



( II M'TER 14 



</ x 



Base 



Base 



5? 



i c 



Base 



S9 



Base 



W 



B ° sc (Individually) 



Base 



W 



F, c 



Base 



Base 



=? 



Base 



So*"' 



Base 



;? 



d" (2) 



(1) is absent if the P, Base chromosome contributed to the P 2 Q 
contained a recessive lethal 

(2) is absent if the P, + chromosome contributed to the P 2 y 
contained a recessive lethal 

figure 14-2. The breeding scheme used in the Base technique. 



in hemizygous X-chromosomc loci, that is, 
X-loci without an allele in the Y chromo- 
some. The males used are wild-type, hav- 
ing all normal characteristics including ovoid, 
dull-red compound eyes. The females have 
X chromosomes homozygous for Bar eye 
( B ) , for apricot eye color ( apr ) , and for 
two paracentric inversions inside the left 
arm. The smaller inversion (fnS) lies inside 
the larger inversion (/// sc s] .sc\ whose left 
point of breakage is designated sc 81 and right. 



sc 8 ) which includes almost the entire left 
arm. "Base' 1 derives its name from #ar, 
apricot, .scute inversion. Stock Base fe- 
males (or males) have narrow-Bar eyes of 
apricot color. The genotype of the Base 
female is written 

sc s1 B InS apr sc 8 sc 8i B InS apr sc s 

A wild-type male is mated with a Base fe- 
male and the F, daughters obtained are 
sc 81 B InS apr sc 8 and appear heterozy- 



Point Mutations 



197 



gous (wide) Bar (and wild-type otherwise). 
Since the very short right arm of the X 
is entirely heterochromatic, it is of no con- 
cern here. Because each F, female is hetero- 
zygous for two paracentric inversions, any 
crossing over between the left arms of her 
X's produces dicentric or acentric crossover 
strands which fail to enter the gametic nu- 
cleus (see Chapter 12). Accordingly. F, 
females produce eggs having an X that is, 
for our purposes, either completely maternal 
(sc 81 B InS apr sc H ) or completely paternal 
( + ) in derivation. If this F, daughter 
mates with her Base brothers, half of the 
sons in the next generation (¥ 2 ) receive the 
+ maternal X, and half receive the Base 
maternal X. So, if the progeny of a single 
Fi female are examined, it is a simple matter 
to detect the presence of both types of sons 
among the more than 80 F 2 progeny usually 
produced. Note that each wild-type F 2 son 
carries an identical copy of the X which the 
mother (the Fi female) received from her 
father (the Pi male). Even when the sperm 
used to form the Fi female carries an X- 
linked recessive lethal mutant, the F: fe- 
male usually survives because she carries 
its -f- allele in her Base chromosome, Each 
wild-type F L > son, however, carries this mu- 
tant in hemizygous condition and usually 
dies before adulthood, so that no wild-type 
sons appear in F>. It becomes clear, then, 
since an Fi female is formed by fertilization 
with a wild-type X-carrying sperm, the ab- 
sence of wild-type sons among her progeny 
is proof that the particular P y sperm carried 
a recessive lethal, X-linked mutant. 

Such a lethal mutant must have occurred 
in the germ line after the fertilization that 
produced the Pi male; he would not have 
survived had it been present at fertilization. 
It is unlikely that many of the X-linked 
lethals detected in sperm originate very early 
in development, for in this case a large por- 
tion of the somatic tissue would also carry 
the lethal and usually cause death before 



adulthood. Even when a few hundred sperm 
from one male are tested, only one is usually 
found to carry an X-linked recessive lethal 
mutant. This indicates that most X-linked 
lethals present in sperm involve only a very 
small portion of the germ line. Occasion- 
ally, however, the mutation occurs early 
enough in the germ line so that several sperm 
tested from the same male carry what proves 
to be the same recessive lethal. 

When a thousand sperm from normal, un- 
treated males are tested for X-linked reces- 
sive lethals by means of a thousand separate 
matings of Fj females, approximately two 
of these matings are found to yield no wild- 
type sons. This X-linked recessive lethal 
mutation frequency of 0.2% is fairly typical 
in D. melanogaster. For every 1000 r of 
X rays to which the adult male is exposed, 
approximately 3.1% more sperm are found 
to carry X-linked recessive lethals (see Fig- 
ure 13-3, for the similar frequency obtained 
after exposure to fast electrons). 

When used as described, the Base tech- 
nique detects only those recessive lethals 
which kill before adulthood. Other reces- 
sive lethals that produce wild-type adult 
males which are sterile or die before they 
can mate are not detected. No recessive 
lethals are detected unless they are hemi- 
zygous in the F 2 male, as mentioned. Since 
a considerable number of X-linked mutants 
whose hemizygous lethality is prevented by 
genes normally present in the Y chromo- 
some is known to occur, this group is 
missed because each F 2 male is normally 
provided with a Y chromosome. Suitable 
modifications of the Base procedure can be 
made to detect this special kind of Y-sup- 
pressed recessive lethal. On the other hand, 
the advantages and applications of the Base 
technique as described are numerous. 

For example, the presence or absence of 
wild-type males in F 2 is easily and objec- 
tively determined. Since the recessive lethal 
detected in F 2 is also carried by the hetero- 



198 



CHAPTER 14 



zygous-Bar F2 females, further study of the 
sive lethal is possible in F L . and subse- 
quent generations. Such studies reveal that 
certain lethals are associated with intergenic 
changes; Lethals not associated with inter- 
genic changes are designated as recessive 
lethal point mutants. The Base technique 
can also be used to deteet recessive lethals 
that occur In a P, Base chromosome, the 
absence of Base males among the F 2 prog- 
eny indicating such a mutation. Moreover, 
if the environmental conditions are standard- 
ized, it becomes possible to detect hemi- 
zygous mutants which either lower the via- 
bility of the F2 males without being lethal 
or raise their viability above normal. The 
opportunity for studying the viability effects 
of recessive lethals in heterozygous condition 
is also provided by this technique. 

Although the Base technique can also be 
used to detect X-linked mutants producing 
a visible morphological change when hemi- 
zygous, all those "visibles" which are also 
hemizygous lethals are missed. The "Maxy" 
technique 5 overcomes this difficulty. In this 
method, the tested female has fifteen X- 
linked recessive point mutants on one homo- 
log and their normal alleles on the other. 

-See H. J. Muller (1954). 



Suitable paracentric inversions maintain the 
identity oi these chromosomes in successive 
generations. Mutants are detected when 
such females show one or more of the re- 
cessive traits. Maxy detects, therefore, any 
mutation involving the normal alleles of the 
fifteen reccssives, provided that the mutant 
does not produce the normal phenotype 
when heterozygous with the recessive allele 
and is not a dominant lethal. Once such 
mutants are obtained, they can be screened 
for point mutants. 

The study of recessive lethals in the X 
chromosome and in the autosomes shows 
that there are hundreds of loci whose point 
mutations may be recessively lethal. It 
should be noted that the recessive lethals 
detected by Base and the visibles detected 
by Maxy are not mutually exclusive types of 
mutants, for some Maxy-detected visibles 
are lethal when hemizygous, and about ten 
per cent of Base-detected hemizygous lethals 
show some morphological effect when hetero- 
zygous. It can be stated, in general, that 
any mutant in homo- or hemizygous condi- 
tion which is a "visible" will produce some 
change in viability, and, conversely, that any 
mutant which affects viability will produce a 
"visible" effect, "visible" at least at the bio- 
chemical level. 



SUMMARY AND CONCLUSIONS 

The mutational units of a genotype are, in order of size: the genome; the chromosome; 
chromosomal segments involving more than one gene; and the gene. Since a recom- 
hinational gene can have multiple alleles, gene mutation may involve the entire recom- 
hinational unit or one or more mutational sites within it. Although the genes delimited 
operationally by recombination and by mutation may not be materially equivalent, we 
shall continue to assume that this is so until we have evidence to the contrary. 

The occurrence of gene mutation is not limited by any ploidy, type of cell or gene, or 
effect it can have on synapsis. It is limited with respect to the effect it can have on 
gene polarity. Tripolar genes are excluded, bipolarity being the usual and unipolarity 
the less usual alternative. 

Point mutations are the remainder of all mutations not identifiable as intergenic 
changes. Since point mutants include gene mutants, the former can be studied to 
reveal the mutational characteristics of the gene. The frequency of point mutants 
increases linearly with the dose of high-energy radiations; there is no effect from dose 



Point Mutations 199 




H. J. Muller, at Cold Spring Har- 
bor, N.Y., 1941. 

protraction and no threshold dose below which the genetic material is safe from change. 
Point mutations also indicate that a given gene is relatively stable over many cell gen- 
erations — changes in genes resulting from very localized physico-chemical events last- 
ing a matter of minutes, after which the new gene is stable. Point mutations are en- 
hanced or induced by temperature changes, aging, gene replication, and physical and 
chemical mutagens. It is possible that changes resulting in point mutants take place 
in the old gene, in the new gene, or during the formation of the new gene. 

Genetic schemes for the detection of X-linked recessive lethal and recessive visible 
mutants in Drosophila are described. A single representative of most normal genes 
fails to produce the full normal phenotypic effect, and most point mutants act on the 
phenotype in a hypomorphic or amorphic manner. The study of point mutants of 
these and other types reveals that almost all are detrimental to the reproductive poten- 
tial of individuals when pure (not hybrid), and to a lesser extent when hybrid. Ac- 
cordingly, most point mutants are not completely recessive to their normal genetic 
alternatives. 



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Alexander, P., "Radiation-Imitating Chemicals," Scient. Amer., 202. No. 1:99-108, 
1960. 

Crow, J. F., and Temin, R. G., "Evidence for Partial Dominance of Recessive Lethal 
Genes in Natural Populations of Drosophila," Amer. Nat., 98:21-33, 1964. 

Muller, H. J., "Variation Due to Change in the Individual Gene," Amer. Nat., 56:32- 
50, 1922. Reprinted in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood 
Cliffs, N.J.: Prentice-Hall, 1959, pp. 104-116. 

Muller, H. J., "Artificial Transmutation of the Gene," Science, 66:84-87, 1927. Re- 
printed in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood Cliffs, N.J.: 
Prentice-Hall, 1959, pp. 149-155, and also in Great Experiments in Biology, 
Gabriel, M. L., and S. Fogel (Eds.), Englewood Cliffs, N.J.: Prentice-Hall, 1955, 
pp. 260-266. 

Muller, H. J., "A Semi-automatic Breeding System ('Maxy') for Finding Sex-linked 
Mutations at Specific 'Visible' Loci," Drosophila Info. Serv., 28:140-141, 1954. 



200 ( HAPTER 14 

\l tiller. H. J.. "The Nature oi the Genetic Effects Produced by Radiation," in Radiation 
Biology, Hollaender, A. (Ed.), Vol. I. Chap. 7:351 473. New York: McGraw- 
Hill, l l >54. 

Mul lor. H. J., and Oster, I. I.. "Some Mutational Techniques in Drosophila," pp. 249- 
278. in Methodology in Basic Genetics, Burdette, W. J. (Ed.). San Francisco: 
Holden-Day, 1963. 

Schalet, A.. "A Study of Spontaneous Visible Mutations in Drosophila Melanogaster," 
Proc. X Intern. Congr., Genetics. Montreal. 2:252 (Abstr.), 1958. 

See Supplement 111. 



QUESTIONS FOR DISCUSSION 

14.1. Is there a safe dose of X rays and/or ultraviolet radiation; that is, a dose that 
cannot produce some point mutations? Explain. 

14.2. Can we he sure that any given mutation involves a single gene change rather 
than intergenic one? Explain. 

14.3. Would we know of the existence of genes if all genes had identical mutational 
capacity? Explain. 

14.4. Would you expect the mutation rate to Polydactyly, P, from normal, /?, to be 
greater among normal individuals in a pedigree for Polydactyly than it is among 
normals in general? Explain. How might you test your hypothesis? 

14.5. Do the mutational properties discussed suggest any limitations with respect to 
the chemical composition of genes? Explain. 

14.6. When a chromosome is broken, is the breaking point within a gene, between 
genes, or both? Justify your answer. 

14.7. Point mutations are sometimes called gene mutations. Do you think this is 
permissible? Why? 

14.8. In what way is the study of mutation dependent upon genes? In what way 
is the reverse true? 

14.9. What is your opinion regarding the validity of applying principles of point 
mutation directly to gene mutation? 

14.10. Are all of the mutants detected by the Base or Maxy techniques point mutants? 
Explain. 

14.11. Suppose, in the Base technique, an F 2 culture produced both of the expected 
t\pes of daughters but no sons at all. To what would you attribute this result? 

14.12. How can you determine whether a recessive lethal detected in the F L > by the 
Base technique is associated with an inversion or a reciprocal translocation? 

14.13. A wild-type female produces 110 daughters but only 51 sons. How can you 
test whether this result is due to the presence, in heterozygous condition, of a 
recessive X-linked lethal? 

14.14. How can you explain the phenotype of a rare female in the Maxy stock that 
produces only unexceptional progeny but has compound eyes distinctly lighter 
than normal? 

14.15. Compare the relative suitability of man and Drosophila for the determination 
of mutation frequencies. 

14.16. The genes in the X chromosomes are incompletely linked in the females of 
the Base stock. Do you agree with this statement? Why? 



Chapter 15 

THE GENE POOL; 
EQUILIBRIUM FACTORS 



T! 



|he recombinational and muta- 
tional properties of the genetic 
material have been studied in 
cross-fertilizing individuals and the nature 
and phenotypic consequences of various ge- 
netic units have been described in terms of 
the traits found in such individuals and their 
relatives. Cross-fertilizing individuals are 
members of a general population. In a 
general population, each individual usually 
has an opportunity to choose a mate from 
a large number of the other members. The 
gametes of all mating individuals furnish a 
pool of genes, or gene pool, from which the 
genes of the next generation are drawn. 
Over successive generations what happens to 
the frequency of a particular gene in the 
gene pool? Let us construct a gene pool 
and investigate this question. 

Suppose that Mars is colonized by human 
beings, that the population sent there is suffi- 
ciently large, and that — with respect to eye 
color genes — only the B (brown) allele and 
the completely recessive b (blue) allele are 
present in the gene pool in the frequencies 
.2 B and .8 b. Presuming that marriages are 
not influenced by eye color phenotype, what 
genotypes and phenotypes will the F, have? 
The answer can be seen in Figure 15-1. 
As the result of the random union of gam- 
etes, 4% of these children are BB; 32% 
are Bb; and 64%, bb. Phenotypically, the 
Fi population is composed of 36% brown- 
and 64% blue-eyed people. 

In the absence of mutation, what is the 

201 



gene pool in the gametes of the F,? The 
4% of F, BB individuals furnish 4% of all 
gametes, and these carry B. The 32% of 
F, Bb individuals furnish 32% of all gametes 
of which half (16% ) carry B and the other 
half b. Therefore, the total gene pool con- 
tains 20% gametes with B. The b gametes 
comprise 80% of the gene pool (16% from 
the 32% of Bb heterozygotes and 64% from 
the 64% of bb individuals). Note that the 
gene pool of the F, is identical to that of the 
P,. Therefore, in the F 2 and all subsequent 
generations, the same genotypic and pheno- 
typic ratios are found, because the fre- 
quencies of B and b in the gene pool remain 
constant. 

What would be the consequence if, in- 
stead of starting the Martian colony with a 
gene pool of 20% B and 80% b, some other 
proportion were used? We can generalize 





FEMALE 


GAMETES 




.2 B 


.8 b 




.04 BB 


.16 Bb 


.2 B 


Brown 


Brown 


MALE 


Eyes 


Eyes 


GAMETES 








16 Bb 


.64 bb 


.8 b 


Brown 


Blue 




Eyes 


Eyes 



The F, Population 

.04 Brown (BB) + .32 Brown (Bb) + .64 Blue (bb) 

The F, Gene Pool 

B = .04 + .16 = .2 
b = .16 + .64 = .8 

figure 15-1. F, genotypes and the gene pool 
these produce. 



202 



CHAPTER 15 



EGGS 
p B (Brown) q b (Blue) 



SPERMS 



P B 

(Brown) 

qb 
(Blue) 



p BB 
(Brown Eyes) 



p q Bb 

(Brown Eyes) 



p q Bb 

(Brown Eyes) 



q 2 bb 
(Blue Eyes) 



FIG1 ki 15-2. The types and frequencies of 
genotypes produced by a gene pool composed 
oj p B nnd q b. 



the analysis by letting p equal the fraction 
of male and female gametes in the popula- 
tion which carries B. and q equal the frac- 
tion which carries /). Naturally, for eggs, 
p -+- q = 1, as is also true for sperm. These 
sex cells combine at random to produce the 
result shown in Figure 15-2. The offspring 
population, then, is 

p- BB + 2 pq Bb + q 2 bb 

The fraction of brown-eyed individuals is 
p- + 2 pq, whereas q- is the blue-eyed frac- 
tion. The frequency of B and b among the 
gametes produced by the offspring popula- 
tion is: 

B = p- + pq = p(p + q) = p 

b = q 2 + pq = q(q + p) = q 

Thus the gene frequencies have remained 
the same as they were in the gametes of the 
previous generation, and all future genera- 
tions will have the same gene pool and the 
same relative frequencies of diploid geno- 
types. The formula 

p- BB + 2 pq Bb + q- bb 

describes the genotypic equilibrium produced 
by a static gene pool. 1 

1 This is called the Hardy-Weinberg equilibrium 
principle. 



It should be noted that this equilibrium 
principle is independent oi the occurrence 
of dominance. Moreover, the B and b in 
the formula can represent any two alleles 
whose frequency in the gene pool is known, 
even if the sum of their frequencies is less 
than one, as in cases of multiple allelism. 

If this equilibrium principle applied in- 
definitely, gene frequencies would remain 
unchanged, and the evolution of different 
genotypes and their resultant new pheno- 
types would not occur. In the Martian 
model described, certain conditions had to 
be fulfilled in order to maintain a genetic 
equilibrium. One condition was met by 
barring mutation, for if it were permitted, 
obviously the frequency of the two alleles, 
B and b, in the population would have 
been reduced, and the equilibrium upset. 
The frequency of any allele would also 
have been changed if the mutation rates to 
and from it were different. In either or both 
types of events, the genetic equilibrium is 
shifted until a new one is attained. There- 
after, the new equilibrium is maintained 
until some new factor acts on mutation rate 
in a directional way. 

Our model also assumed that the repro- 
ductive potential ( biological fitness, or adap- 
tive value) was the same regardless of the 
genotype for eye color. But it is possible, 
under certain conditions, that persons with 
blue (or with brown) eyes are preferred as 
mates, in which case the reproductive po- 
tential of an individual is not independent 
of the alleles under consideration. Accord- 
ingly, if individuals with a certain genetic 
endowment produce more surviving off- 
spring than those produced by a different 
genetic endowment, the genes which transfer 
this higher biological fitness tend to increase 
their frequency in the population, whereas 
those genes with lower fitness tend to de- 
crease it. In this way selection, by operat- 
ing on genotypes of different adaptive value. 



The Gene Pool; Equilibrium Factors 



203 



causes changes in gene frequencies and shifts 
in the genotypic frequencies found at equi- 
librium. 

The Martian population was also pre- 
sumed to be large. Suppose, however, that 
the Martian population (whose gene pool is 
2B and .8b) ran short of food, and only 
one couple, determined by chance, could 
have children. The chance that this hus- 
band and wife would be blue-eyed is 
.64 X .64, or about .41. Accordingly, there 
is a 41% chance that the gene pool will 
drift at random in this particular manner, 
producing the new gene frequencies of 1 .0 
for b and for B. This random genetic drift 
can also be illustrated in a less extreme 
situation: If a population is very large, and 
a certain family happens to produce a rela- 
tively large number of children for several 
generations, then the proportion of all indi- 
viduals in the population with this family 
name is still very small. But if the popula- 
tion decreases while this family's reproduc- 
tive rate is unchanged, the proportion of the 
population with this surname increases. Ac- 
cordingly, when populations are very large, 
oscillations in the number of children pro- 
duced by different genotypes occurring by 
chance are unimportant, for they do not 
change the gene pool. In small populations, 
however, such chance oscillations can change 
gene frequencies via random genetic drift. 

In our Martian model, the possibility that 
the colony would have emigrants or immi- 
grants was not considered. If the emigrants' 
gene frequencies are different from those re- 
maining in the population gene pool, then 
the gene frequencies in the remaining popu- 
lation will be changed. If the immigrants' 
gene frequency is different from the natives', 
and they interbreed, the gene pool will again 
be changed. In this way migration can shift 
the genetic equilibrium. 

We see then that a cross-fertilizing pop- 
ulation remains static — in genetic equilib- 



rium — in the absence of mutation, selection, 
random genetic drift, and differential migra- 
tion. The occurrence of one or another or 
all of these factors changes the frequencies 
of genes in the gene pool and thereby shifts 
the frequencies of genotypes at equilibrium. 
Different species possess different gene pools, 
and it is natural to presume that they are 
different species because of their different 
gene pools. Accordingly, the factors which 
change gene frequencies are considered to 
be the main causes of species formation. 
Insofar as the formation of higher taxonomic 
categories is, like speciation, based upon 
change in gene pools, the principal causes 
of biological evolution are: 

1. Mutation (which supplies the raw ma- 
terials) 

2. Selection (which shapes these raw ma- 
terials into the biologically fit geno- 
types of races and species) 

3. Random genetic drift (which can pro- 
duce rapid changes in gene frequency 
in small populations) 

4. Differential migration (which can shift 
gene frequencies via interchange of in- 
dividuals between populations). 

* Selection of Genotypes 
The disequilibrating effect of selection upon 
the gene pool has already been noted. Selec- 
tion acts at the phenotypic level to conserve 
in the population those genotypes which pro- 
vide the greatest reproductive potential. Se- 
lection takes place at all stages in the life 
cycle of an individual. Since it acts to pre- 
serve whole phenotypes and not single traits, 
selection conserves genotypes and not sing'e 
genes. Sometimes selection acts upon the 
phenotypes produced by single genomes in 
haploid species or stages; at other times — 
in sexually reproducing organisms — it acts 
upon the combined phenotypic effect of two 
genomes. It should be noted that what is 
a relatively adaptive genotype at one stage of 



204 



CHAPTER 15 



the life cycle may be relativelj ill-adaptive at 

another whether or not these stages have 
the same or different ploidies. It is. of 

course, the total adaptiveness of all these 
separate features which determines the over- 
all reproductive potential oi an individual. 
Finally, it should he noted that in cross- 
fertilizing populations, selection favors geno- 
types which produce maximal fitness of the 
population as a whole. Because selection 
acts this way, it is possible that some por- 
tion of the population receives genotypes 
which are decidedly not advantageous. If 
this is so. the same genetic components are 
expected to be advantageous when present 
in en her. more probable, combinations. 

* Selection Against Mutants 

Since the human being is primarily a diploid, 
it is upon the diploid-produced phenotype 
that selection principally operates. If one 
asks, "What is the fate of mutants in the 
gene pool?*' the answer must include knowl- 
edge of the frequency with which the mu- 
tants arise as well as their effects upon re- 
productive potential in a diploid genotype. 
Remember that the phenotypic effect of a 
mutant gene depends not only upon the na- 
ture of its allele but also upon its relationship 
to the rest of the genotype. 

Let us consider, in turn, the fate in the 
human gene pool of mutants whose overall 
phenotypic effect is: dominant lethal; domi- 
nant detrimental; recessive lethal; or reces- 
sive detrimental, as influenced by selection 
and mutation. 

Dominant lethal mutants are lethal when 
heterozygous and are eliminated from the 
gene pool the same generation they arise. 
Accordingly, the biological fitness of such 
mutants is zero. If the normal homozygote 
(A i A ] ) is considered to have a selective 
disadvantage of zero, then the dominant 
lethal is at a complete selective disadvantage, 
and the selection coefficient, s, is one. We 
can readily see that the imitation frequency, 



ii, of this dominant lethal condition must 
equal one half the frequency of affected in- 
dividuals (A1A2), since each affected indi- 
vidual has one mutant and one normal gene. 
In the absence of special medical treatment. 
retinoblastoma, a type of cancer of the eye, 
is an example of such a dominant lethal in 
man. 

A chondroplastic ( or chondrodystrophy ) 
dwarfism is characterized by disproportion 
— normal head and trunk size but shortened 
arms and legs. This rare, fully penetrant 
(see p. 73) disease is attributed to the pres- 
ence of a gene in heterozygous condition 
which therefore acts as a dominant detri- 
mental mutant. Such dwarfs (A^A 2 ) are 
known to produce only 20 per cent as many 
children as normal people. Because of this 
lower reproductive potential the A\A-> geno- 
type selection coefficient is .8. 

In one study the frequency of A x A- 2 in 
the population was found to be 10 dwarf 
babies in 94,075 births. The dwarf children 
in this sample must have resulted from nor- 
mal parents who carried new mutations to 
A ■> or from one normal and one dwarf par- 
ent. The gene frequency, p, of A-< in the 
population, therefore, must be 10 per 
2(94,075) or .000053. From the incidence 
of dwarfs known to have normal parents the 
mutation frequency, u, to A 2 is .000042. If 
the value s = .8 is correct, then p = u/s, or 
.000042 .8, or .0000525, which is in excel- 
lent agreement with the gene frequency (p) 
value observed. Gene frequency for a 
dominant lethal equals mutation frequency 
(p = u) because s = 1; however, in the 
present case s is less than one, so the gene 
frequency is greater than the mutation fre- 
quency. Actually the gene frequency for 
dwarfism is not very much larger than the 
mutation frequency, demonstrating the effi- 
ciency of natural selection in eliminating 
such mutants from the gene pool. 

The gene for juvenile amaurotic idiocy 
{a-,) has no apparent effect when hetero- 



The Gene Pool; Equilibrium Factors 



203 



zygous (A^a-j)', since homozygous children 
die, a-2 is a recessive lethal mutant. Affected 
individuals are found with a frequency of 
one per 100,000, or .00001. What is the 
frequency of a 2 in the gene pool? As shown 
in Figure 15-3, the frequency of a^a% indi- 
viduals at equilibrium is equal to q-. Ac- 
cordingly, the frequency of a- 2 (q) must be 
equal to Vq 12 , or V .00001, or about .003, 
whereas the frequency of A\ must be one 
minus .003. or .997. Note that heterozy- 
gotes (carriers) are 600 times more fre- 
quent than afflicted homozygotes. What is 
the mutation frequency from A x to a-P. As- 
sume that the gene pool is at equilibrium; 
in other words, the frequency with which 
a> enters the population by mutation equals 
the frequency with which it leaves the pop- 
ulation in flotfo homozygotes. Accordingly, 
the mutation frequency to a 2 must be .00001. 
The selection coefficient for normal indi- 
viduals (A X A X and A^a^) is zero, and for 
a 2 a-2 it is one. We see, therefore, that at 
equilibrium the frequency of a recessive 



GENOTYPE 
PHENOTYPE 



FREQUENCY AT 
EQUILIBRIUM 



A, A, 



A,a : 



Normal Normal Dies 
P 5 2pq q* 



u = Mutation rate from A, — a, 

q ="U u/s Here s = 1, hence q ="\| 

u = 10~ 5 = 0.000,01 Hence q =M 0.000,01 = 0.003 



ACTUAL 

FREQUENCY 

AT 


<p 2 > 
(0.997)' 


(2pq) 

2I0.997M0.003) 


tq 2 ) 
(.003I 2 


EQUILIBRIUM 


0.994 


0.006 


0.000,01 




OtQ 



DtO 



(WSSd 



lolorioo 



figure 15-3. Juvenile amaurotic idiocy. {See 
text for explanation.) 



figure 15—4. Pedigree showing the occur- 
rence of phenylketonuria among the offspring 
of cousin marriages {denoted by thick marriage 
lines). 



mutant in the gene pool can be expressed as 
q = Vii s, where s = 1 for a recessive 
lethal. When the homozygous recessive 
mutant is detrimental without being lethal, 
s becomes less than one (but more than 
zero) and the frequency of the mutant in 
the gene pool increases. Thus, if s were 
Y 4 instead of one, q would be twice as large. 

* Nonrandom Matings 

In deriving the types and frequencies of 
genotypes in a population at equilibrium, 
we assumed that marriages were random 
with respect to the genotypically-determined 
trait under consideration. Such a randomly- 
mating population is said to be panmictic 
or to undergo panmixis. What happens if 
the different genotypes do not marry at ran- 
dom? Consider the disease phenylketonuria 
(Figure 15-4) which involves a type of 
feeblemindedness in individuals who are 
homozygous for a recessive gene, and who 
metabolize the amino acid phenylalanine to 
phenylpyruvic acid, which is toxic. The 
frequency in the gene pool of the normal 
gene (A) is .99 and of the abnormal gene 



206 



CHAPTER 15 



(a) is .01. In the population at equilib- 
rium, therefore. AA:Aa:aa individuals have 
frequencies oi 9801 10,000: 198 10,000: 
I 10,000, respectively. Notice that Aa in- 
dividuals are 1 C )S times more frequent than 
aa, so that even if every aa did not repro- 
duce, only one per cent of the a genes pre- 
sent in the gene pool would be eliminated 
each generation. This fact illustrates the 
inefficiency of selection against homozygotes 
for rare recessive genes, at least insofar as 
lowering the frequency of such genes is con- 
cerned. A A and Aa individuals apparently 
marry at random but feebleminded people 
do not. So panmixis does not occur with 
respect to this trait, and persons with dif- 
ferent genotypes tend to be restricted in their 
marriages — all the available marriage part- 
ners making up a person's reproductive 
isolate. The occurrence of different repro- 
ductive isolates for normals and phenyl- 
ketonurics has little effect on the relative 
frequencies of different genotypes in succes- 
sive generations, because aa people have so 
few of all the a genes present in the popula- 
tion. Clearly, only marriages between two 
A a individuals are of consequence, since 
those are the major source of aa offspring. 

The example of phenylketonuria shows 
that when a gene is rare and apparently com- 
pletely recessive, nonrandom marriage has 
little influence either upon gene frequency 
or the diploid (heterozygous or homozygous) 
genotypes in which it is found in the popula- 
tion. When the mutant is relatively fre- 
quent in the population, however, it is ob- 
vious that nonrandom marriages raise the 
frequencies of certain diploid genotypes and 
lower others. Moreover, if there are adap- 
tive differences for the different genotypes, 
the composition of the gene pool can be 
changed in a different direction or at a dif- 
ferent frequency than would be predicted 
for a population mating at random. 

Consider two ways in which mating can 



be nonrandom. The first invokes the tend- 
ency of phenotypically similar individuals 
(except for sex) to mate and is referred to 
as assortive mating. This kind of breeding 
pattern is generally true in animals including 
human beings. The genetic result is the 
production of more homozygotes than would 
occur by randomly-chosen matings. 

The second departure from random mat- 
ing involves inbreeding, the tendency for 
mates to be more closely related in descent 
than randomly chosen mates. What is the 
effect of inbreeding carried out for a single 
generation? This can be determined by 
studying what happens to genes that are 
heterozygous in the parent generation. 
There are various degrees of inbreeding, the 
closest form being self-fertilization. In self- 
fertilization the heterozygote for a given pair 
of genes, A a, produces progeny of which one 
half are homozygous. In general, the de- 
crease in heterozygosity because of self- 
fertilization can be expressed as follows: 
the chance that an offspring receives a given 
gene in the male gamete is y 2 , and the 
chance that it receives the same allele in 
the female gamete is H; the chance that 
the offspring is a homozygote for that allele, 
therefore, is %. But there is an equal 
chance that the offspring becomes homozy- 
gous for the other allele, so that the total 
chance for homozygosis from this type of 
inbreeding is 50%. If all members of the 
population are heterozygotes and self-fer- 
tilize, then in each successive generation, 
half of the genes that were heterozygous 
become homozygous. 

Suppose, on the other hand, that a portion 
of a population mating at random has X% 
homozygous individuals. These could come 
from matings between two heterozygotes, 
two homozygotes, or a heterozygote and a 
homozygote. If the gene pool is at equi- 
librium, the random matings that tend to 
increase homozygosis are counterbalanced 



The Gene Pool; Equilibrium Factors 



207 



by others which decrease it, so that X% 
homozygosis remains constant generation 
after generation. Consider what happens in 
another portion of this population which 
happens to practice self-fertilization for one 
generation. Since this segment of the pop- 
ulation already shows X% homozygosis, its 
offspring will also have X% homozygosis. 
But, if this segment is Z% heterozygous, 
after self-fertilization the offspring will have 
only V-2 Z% heterozygosis, and, therefore, 
will show a total homozygosis of X% + % 
Z% . In other words, each generation of self- 
fertilization makes half of all heterozygous 
genes homozygous, and, in a normally ran- 
dom-mating population, the effect of self- 
fertilization is to increase the random-mating 
frequency of homozygosis by y 2 the fre- 
quency of heterozygosis. 

How much is homozygosity increased in 
brother-sister (sib) matings? The chance 
that a particular gene in the father is present 
in the male sib is y 2 , and the chance that 
the male sib's child receives this is similarly 
] {>; the chance for the occurrence of both 
events is %. The chance that the female 
sib receives and transmits this same gene to 
her child is also %. Therefore, the chance 
that the child of the sib mating receives two 
representatives of this same allele is l / 4 times 
% j or it has ] \ 6 chance of being homozygous 
for this gene. Since the child has an equal 
chance to become a homozygote for the 
other allele in his grandfather and for each 
of the two alleles in his grandmother, this 
gives him 4 times y 16 or a 25% chance of 
homozygosis. In other words, sib matings 
cause % of the heterozygous genes to be- 
come homozygous. This chance of homo- 
zygosis from sib mating is in addition to 
the chance of homozygosis from mating at 
random. 

Matings between individuals who have 
one parent in common are called half-sib 
matings. In this case, the frequency with 



which a given allele in the common parent 
passes to the male half-sib is y 2 , and the 
frequency with which an offspring of this 
sib receives this allele is l / 2 ', the chance of 
both events occurring is, therefore, %. The 
chance is also r 4 for these events to occur 
through the female half-sib, so that the 
chance of a given allele becoming homo- 
zygous from a half-sib mating is y 4 times 
y 4 , or y 16 . Since the other allele in the 
common parent could, in this way, also be- 
come homozygous y 16 of the time, the com- 
bined additional chance of homozygosity for 
half-sib matings is 1 £, or, in other words, 
y 8 of the heterozygous genes become homo- 
zygous because of this type of inbreeding. 

The amount by which heterozygosity is 
reduced because of inbreeding is called the 
inbreeding coefficient, F. In a similar man- 
ner we can determine that in the case of 
cousin marriage, F is V lr> . The values of F 
for more complicated pedigrees can be 
worked out accordingly. 

All forms of inbreeding increase homozy- 
gosity. Let us calculate the consequence of 
cousin marriage upon the frequency of 
phenylketonuria. Its frequency of hetero- 
zygotes per 10,000 people is 198 (see p. 
206). Cousin marriage reduces heterozy- 
gosity by y 16 , or by twelve individuals, of 
which half of them are expected to be nor- 
mal (A A) and half affected {ad). Since 
random mating produces one affected indi- 
vidual per 10,000, cousin marriages bring 
the total number of affected homozygotes 
in this population to seven (six from in- 
breeding, one from random breeding). Ac- 
cordingly, there is a sevenfold greater chance 
for phenylketonuric children from cousin 
marriages than from marriages between un- 
related parents. 

Another example of how cousin marriages 
increase the risk of defect comes from a 
study which found that in a Japanese pop- 
ulation (Figure 15-5) congenital malforma- 



208 



CHAPTER 15 



nons. stillbirths, and infant deaths were 24 
to 48 per cent higher when cousins married 
than when parents were unrelated. Since, in 
sonic cases, delects such as these are known 
to be due to recessive genes in homozygous 
condition, these results support the view that 
homozygosis resulting from inbreeding can 
produce detrimental effects. Although in- 
breeding produces homozygosis and homo- 
zygosis can lead to the appearance of de- 
tects, it must not be inferred that inbreeding 
is disadvantageous under all circumstances. 
Many individuals do become homozygous 
for detrimental genes as a result of inbreed- 
ing, but just as many become homozygous 
for the normal alleles. The success of self- 
fertilizing species is testimony to the advan- 
tage of homozygosity at least for some types 
of organisms. 

* Heterosis 

In normally cross-fertilizing species, how- 
ever, inbreeding usually results in a loss of 
vigor which is directly linked to homozygosis. 
What is the functional basis for the adaptive 
superiority of heterozygotes. usually known 
as heterosis or hybrid vigor? Consider the 
three genotypic alternatives, A A, A A', A' A' 
relative to their phenotypic effects. Suppose 
A' A' is less vigorous than A A. Whether A 



is completelj or incompletely dominant to A' 

or shows no dominance to it. the A A' hetcr- 
OZygOte will be superior to one of the 
homozygotes. It is also possible that the 
heterozygote has a greater adaptive value 
than either type of homozygote. To illus- 
trate this possibility, imagine that A is 
pleiotropic. having a relatively great adaptive 
effect with respect to trait X but a relatively 
less adaptive effect with respect to trait Y. 
whereas the reverse is true of A', namely, 
relatively less adaptive for X and relatively 
more adaptive for Y. In the event of no 
dominance, the heterozygote is superior to 
either homozygote. Heterosis can be pro- 
duced, therefore, when the heterozygote is 
superior to either one or both homozygotes. 
The first type of heterotic effect can be 
demonstrated by crossing two pure lines, 
homozygous for different detrimental reces- 
sives (A A bb CC dd by aa bbCC DD). 
The F, (Aa bb CC Dd) is uniform yet more 
vigorous (having normal alleles at three loci) 
than either parent (each of which had nor- 
mal alleles at two loci) because the domi- 
nant alleles hide the detrimental effects of 
the recessive ones. In this case the hetero- 
zygous F 2 progeny carrying Aa bb CC Dd 
are no more adaptive than the homozygotes, 
AA bb CC DD. 



Frequency from Increase in Frequency Per cent 

Unrelated Parents with Cousin Marriage Increase 



CONGENITAL 
MALFORMATION 



.011 



.005 



48 



STILLBIRTHS 



.025 



.006 



24 



INFANT DEATHS 



.023 



.008 



figure 15-5. Increased risk of genetic defect with cousin marriages. 
Hiroshima and Nagasaki. ) 



34 

( Data from 



The Gene Pool; Equilibrium Factors 



209 



The second type of heterotic effect can 
be illustrated in human beings. As men- 
tioned on p. 71, homozygotes for the gene 
for sickle cell anemia (/3 B [J a ) usually die 
from anemia before adolescence. f} A {l A in- 
dividuals have normal blood type, whereas 
ff A fj a individuals are either normal or have 
a slight anemia. In certain countries the 
frequency of /3 s in the gene pool follows the 
expectation for a recessive lethal gene. In 
other countries, however, fj s is more fre- 
quent than expected. This difference is at- 
tributable to the (3 A f3 s heterozygote being 
more resistant to certain kinds of malaria 
than the f3 A f3 A homozygote. Of course, in 
nonmalarial countries, (3 s confers no anti- 
malarial advantage, and so the fitness of the 
heterozygote (1 — s) is lower than that of 
the normal homozygote ( 1 ) , whereas the 
,3' s 7? 8 individual has a fitness of zero. As 
expected, therefore, sickle cell anemia is rare 
or absent in most of the world where certain 
forms of malaria are absent. 

On the other hand, in certain malarial 
countries, even though heterozygotes may be 
slightly anemic, the advantage of being re- 
sistant to malaria produces a greater overall 
fitness than does the (3 A /3 A genotype. Here 
the fitness of the heterozygote, f3 A (3 s , is 
maximal and therefore must be assigned the 
value one, whereas that of the normal homo- 
zygote, f3 A p A , is one minus Si. Mutant 
homozygotes, /? >s '/?' s ', have a fitness of one 
minus s 2 , where s 2 equals one, since all f3 s (3 s 
die (even if extremely resistant to malaria). 
In this situation natural selection maintains 
both fi A and /3 s in the gene pool, (3 s having 



a frequency equal to 



This fraction 



Si + s 2 

can be read as "the advantage of f3 s (as 
shown by the advantage of fl A (3 s over /3 A f3 A ) 
divided by the total disadvantage of f3 A and 
/3 s ." Thus, when the heterozygote, being 
more adaptive than either homozygote, 
shows heterosis in this way, natural selection 



maintains a gene such as (1* in the gene pool 
even though it is lethal when homozygous. 

Although we have discussed heterosis in 
terms of the phenotypic effects of the mem- 
bers of a pair of alleles, it should not be in- 
ferred that the unit of heterotic action is 
always a single pair of genes. Since we 
know that different pairs of genes interact 
in various ways to produce phenotypes, it 
would not be surprising to find that heterosis 
results from the effects of combinations of 
nonalleles and alleles. 

Natural populations of Drosophila pseudo- 
obscura contain various paracentric inver- 
sions. Laboratory populations can be 
started with some individuals carrying the 
normal chromosome arrangement and others, 
a particular one of these inversions. After 
a number of generations has passed, in some 
cases the population comes to contain only 
normal chromosomes, because the inversion 
chromosome behaves like a detrimental gene 
which provides no advantage when hetero- 
zygous and is eliminated from the gene pool. 
When other particular inversions are tested 
this way, however, an equilibrium is reached 
— both the normal and inverted chromo- 
somes are retained in the gene pool. In 
these cases, the inversion heterozygote is 
adaptively superior to either homozygote, 
showing heterosis just as the gene for sickling 
in malarial countries. It is difficult to decide 
the genetic basis for heterosis in such cases, 
however, since the hybrid vigor could be 
due to: the genes gained or lost at the time 
the inversion was initially produced; or the 
new arrangement of the inverted genes; or 
the types of genes or groups of genes con- 
tained within the inversion. Recall that in- 
dividuals with paracentric inversions are not 
at a reproductive disadvantage in Drosophila 
and suppose a heterotic system exists or de- 
velops in Drosophila heterozygous for a 
paracentric inversion. If the heterosis is due 
to the action of several specific nonalleles 



210 



CHAPTER 15 




figure 15-6. The variability of normal corn is pointed out by James F. Crow. 
{Photographed in 1959 by The Calvin Company.) 



within the inverted region, this adaptively 
favorable gene content tends to remain in- 
tact in the heterozygote because of the fail- 
ure of single crossovers within the inverted 
region to enter the haploid egg nucleus. 

Breeding procedures that result in hybrid 
vigor have been widely applied to econom- 
ically important plants and animals. For 
example, it has been estimated that the use 
of hybrid corn has enriched society by more 
than a billion dollars. We might ask: What 
is wrong with normal corn? The answer is 
that it is too variable in quality and vigor 
(Figure 15-6). Inbreeding decreases vari- 
ability, but unfortunately inbreeding also re- 
sults in loss of vigor or other desirable 
traits. The way to overcome this problem 



is to obtain inbred lines which are uniform 
(because they are homozygous) and carry 
different favorable dominant genes (yet are 
also homozygous for various undesirable re- 
cessive genes ) , and cross the different inbred 
lines to each other. Their Fi will be multi- 
ply-heterozygous, uniform, and more vigor- 
ous than cither parental inbred line. 

Consequently, hybrids are made from two 
selected inbred lines — of corn in this case. 
Although the Fi plants are vigorous and 
uniform, they come from kernels grown on 
one of the less vigorous inbred lines. For 
this reason, hybrid seeds are not sufficiently 
numerous, and consequently, commercially 
unfeasible. In practice this difficulty is over- 
come (Figure 15-7) by crossing four se- 



The Gene Pool; Equilibrium Factors 



211 



INBRED A INBRED B 



INBRED C INBRED D 




SINGLE CROSS 
CxD 



SINGLE CROSS 
AxB 



ffll J DOUBLE CRO 

yjri (AxB)x(CxD 



CROSS 

) 




figure 15-7. The production of commercial hybrid 
corn by the "double cross" breeding procedure. 



212 CHAPTER IS 

■ inbred lines img sold inexpensively. Hel 

practical importance; a fuller understanding 
single cross fu brids arc then this phenomenon requii 

other Since I rous the biochemical, molecular I c % c I . 

rbrid plant, seeds produced b) - . and w 

touble Ct plentiful .\n<.\ can be 



SUMMARY AND CONCLUSIONS 

I he whose functi produce the I 

popi* pool The gene pool and the l 

•n> will remain forever unchanged if: the population 
rtk drift does not occur, mutation does not - 
direction preferentially . no differential selection is made I 

•IK |iist like the • lr however. conditions is not 

Bed. a shift in the composition of the gene pool will occur, in other words, 
frequencies will change and so will the frequencies of different genotvpes until a new 
equilibrium is attained. 

It 't onlv species formation hut all oH hiological evolution is b 

upon changes in the gene pool 

I he roles th..r mutation and selection have in establishk mc equilibrium is 

discussed tor those rare mutants which lower reprodlK 
lethal, dominant detriment.il. recessive lethal, or l lemmental 

lorn breeding resulting the 

freq' I he per generation rate oi reduction in betel 

due to inbreeding is ' : tor self-fertilization, '» for sib • »r half-sib 

and ¥u for cousin matings. H<" Lilizing individuals leads 

i h\ heterosis, or h\bnd '• 

Hel • i phenotvpic result ol in because the heiero- 

idaptive! ft to one or to both types o\ homo' peal tm 

nee economically. 



REFERENCES 

\lli». klc Cells and Evolution." Scical tin >5:87 M 

If. polls H' 

and the Origin of Sp )rd EdL, New Yofi Cob 

; ' 
Doha netics and M in WHej 

Gowcn, I. W. (Ed), A wa: low . S 

Mixed Populatioi • 

Reprinted a < Peters, J. A. (Ed.). Eng -wood 

, s i Pi H trimenta in 8 

,,! m i od ( tiffs N I Prentice-Hall, 

■ •.v-Hiil. 



I in- (icnr Pool, Equilibrium i<i<i<>i\ 



•i ; 



Rasmuson m Genetics on iii< Population Level, Stockholm, Sweden Svenska ii"i 
f( n I a gel Bonnici i; i ondon Heincmann 1 961 

Spi.s, I H (Ed.) Papers on \nimal Population Genetics, Boston l ittle, Brown 
196 ■ 

■. ii i i . .Lin i V., Kessingei . M . A and Harris, w Differential Rates ol Di vclopmenl 

ol I leteroth .mil Nonheterotii V< g Mai/c Seedlings I < orrelation ol Dill 

ii.il Morphological Development with Physiological i>iii In Qermlnating 

s,. ,i. I'm.. Nal A. id. Sci U.S 11:212 118 1964 

Spragui ' > I 1 1 ii ) Corn and Corn Improvement, New York \cademii Press \'> • ■ 

Weinberg, W I bei d des Vererbung belm Menschen lahresh Vercin 

i vaterl Naturl In WUrttemberg 64 168 182 1908 rranslated In part, In 
Stern ( i he Hardy Weinberg I • I len >7: 137—138, 194 I 



'jiir.flONS FOR DISCUSSION 

i I i An- the ' iiises "i evolution the lame In populations reproducing onl) asexuall) 
.is in those reprodu< ing lexually? i (plain 

i . • Suppose in a population obeying the Hard) Weinberg rul< mutation o 

i.ii mil • 'i iration and changes the compositii i the gene pool H ■ 

man) additional generations are required befori a new gjenotli equilibrium is 
. itabli ihi 'i ' i iplain 



I Ml OD( - III ■ I lOR/.HAI SK ■ iii 1937 




SBWAL] WatOHl U noted fOf his n\r,inli in 

physiological genetics and In the mathematics 
o) population genetics Photograph was taken 

in 193 I 




-It CHAPTER 15 

15.3. Discuss the statement: "The Hardy-Weinberg Law is the cornerstone of evolu- 
tionary genetics." 

15.4. Assuming that the Hard) Weinberg principle applies, what is the frequency 
o\ the gene R h us onlj allele R' is homozygous in the following percentages 
of the population: 49%? 49? ? 25%? 36%? 

15.5. In the United Stales about 709? of the population gets a hitter taste from the 
drug phenyl thiocarbamide (PTC). I hese people are called "tasters" and the 
remaining 309? who get no hitter taste from PTC are called "nontasters." All 
marriages between nontasters produce all nontaster offspring. Every experi- 
mental result supports the view that: a single pair of nonsex-linked genes de- 
termines the difference between tasters and nontasters; dominance is complete 
between the only two kinds of alleles that occur; penetrance o\ the dominant 
allele is complete. 

( a ) Which of the two alleles is the dominant one? 

(b) What proportion of all marriages hetween tasters and nontasters have no 
chance (barring mutation) of producing a nontaster child? 

(c) What proportion of all marriages occurs between two nontasters? Two 
tasters? 

15.6. The proportion of A A individuals in a large crossbreeding population is .09. 
Assuming all genotypes with respect to this locus have the same reproductive 
potential, what proportion of the population should be heterozygous for A) 

15.7. What do you suppose would happen to a population whose gene pool obeyed 
the Hardy-Weinberg rule for a very large number of generations? Why? 

15.8. Can a population obey the Hardy-Weinberg rule for one gene pair but not 
for another? Explain. 

15.9. Explain whether the mutation frequency to a particular allele is of primary im- 
portance in shifting its frequency in the population, when this gene is: 

(a) a dominant lethal in early developmental stages 

(b) a recessive lethal 

(c) phenotypically expressed only after the reproductive period of the individual 

(d) very rare 

(e) present in small cross-fertilizing populations 

15.10. Can the adaptive value of the same gene (15.9) differ in: 

(a) haploids and diploids? 

(b) males and females? 

(c) two diploid cells of the same organism? 

Explain your answer in each case. 

15.1 1. Other things being equal, what will happen to the frequency in the gene pool of 
a dominant mutant whose selection coefficient changes from one to V? If the 
mutant is completely recessive? 

15.12. If persons carrying detrimental mutants never marry, these particular genes are 
removed from the gene pool. Under what conditions is the failure to marry 
likely to appreciably reduce the frequency of detrimental mutants in the gene 
pool? 

15.13. Are inbreeding and assortive mating mutually exclusive departures from random 
mating (panmixis)? Explain. 

15.14. Explain why the inbreeding coefficient, F, is Vie for cousin marriages. 



The Gene Pool; Equilibrium Factors 215 

15.15. Suppose the frequencies of A and a are .3 and .7, respectively, in a population 
obeying the Hardy-Weinberg rule and mating at random: 

(a) What per cent of the population is composed of homozygotes with respect 
to these genes? 

(b) What would be your answer to (a) after one generation of mating hybrids 
only with hybrids? 

(c) How would the conditions in (b) affect the composition of the gene pool? 

15.16. Discuss, from a genetic standpoint, the advantages and disadvantages of cousin 
marriages in man. 

15.17. In Thailand, heterozygotes for a mutant gene that results in the formation of 
hemoglobin E are more frequent in the population than would be expected from 
the Hardy-Weinberg rule. How can you explain this? 

15.18. Two inbred strains of mice and their V l hybrids are tested for locomotor activity 
(measured for each subject in each group during three consecutive five-minute 
periods) and for oxygen consumption. In both these respects the F, hybrid is 
less variable than the parental strains. Propose a genetic hypothesis to explain 
these results. 

15.19. Compare the reproductive isolates of people who were marrying in 1900 with 
those marrying today. Which factors are the same and which are different? Is 
the change desirable from a biological standpoint? Explain. 



Chapter *16 

GENETIC LOADS AND 
THEIR POPULATION EFFECTS 



Genetic Loads in Drosophila 

The fruit fly, Drosophila pseudoobscura, is 
commonly found in northern Mexico and 
the western United States. When collected 
in the wild, almost all its individuals are 
phenotypically alike, except for the sex 
differences, appearing wild-type or normal. 
We cannot accept this phenotypic uniform- 
ity as evidence of genotypic uniformity, how- 
ever, since a Drosophila population appear- 
ing wild-type can conceal considerable 
genetic variability in the form of isoalleles, 
recessive point mutants, reciprocal transloca- 
tions, paracentric inversions, and so on. We 
would like to estimate the genetic load — the 
total amount of this genetic variability pres- 
ent in a natural population of D. pseudo- 
obscura. 1 

D. pseudoobscura has five pairs of chro- 
mosomes — the usual X and Y sex chromo- 
somes, three pairs of large rod-shaped auto- 
somes (II, III, IV), and a dotlike pair of 
autosomes (V) (Figure 16-1). Numerous 
laboratory strains of this species are avail- 
able whose autosomes are marked by various 
point and rearrangement mutants. We can, 
therefore, make a suitable series of crosses 
between laboratory strains and flies collected 
in the wild which will yield information on 
the presence of autosomal mutants in the 
wild-type flies. In practice, autosomes II, 
III, and IV of individual wild-type flies are 
made homozygous to detect the presence of 



the following recessive mutants (see Figure 
6-2): 

1. Lethal (causing death to all individuals 
before adulthood) or semilethal (caus- 
ing more than ninety and less than one 
hundred per cent mortality before 
adulthood) 

2. Subvital (causing significantly less than 
normal but greater than ten per cent 
survival to adulthood ) 

3. Female sterile (sterile to females) 

4. Male sterile (sterile to males). 

The results of this study are summarized 
in Figure 16-2. About 25% of all auto- 
somes tested this way carry a recessive lethal 
or semilethal mutant. Recessive subvital 
mutants are found in about 40% of III chro- 
mosomes tested and in more than 90% of 
II's and IV's tested; mutants causing sterility 
are present in 4 to 14% of tested chromo- 
somes. Obviously the natural population 
carries a tremendous load of detrimental 
mutants. 

How is this load of mutants distributed 
in the fly population? Consider first one 
pair of the autosomes tested. Each member 
has a 25% chance of carrying a lethal or 
semilethal and a 75% chance of being free 
of such mutants. The chance that both 
members of a pair of chromosomes will 
carry a lethal or semilethal is (0.25 )- or 
6.25% . From the data presented we cannot 
tell whether all the lethals and semilethals 
found in a particular pair of autosomes are 



1 The following is based upon 
Dobzhansky and collaborators. 

216 



work of Th. 




FIGURE 16-1. 
Chromosomal 
complement of 
D. pseudoobscura. 



Genetic Loads and Their Population Effects 



217 



MUTANT TYPE PER CENT OF CHROMOSOMES 

II III IV 

Lethal or Semilethal 25 25 26 

Sobvital 93 41 95 

Female Sterile 11 14 4 

Male Sterile 8 11 12 

figure 16-2. Genetic load in natural popula- 
tions of D. pseudoobscura. {After Th. Dob- 
zhansky.) 



allelic (in which case up to 6.25% of zy- 
gotes in nature would be mutant homozy- 
gotes and fail to become adults), or whether 
all the mutants involve different loci (in 
which case 6.25% of zygotes would be hy- 
brid for linked mutants of this kind), or 
whether some combination of these alterna- 
tives is obtained. In any case, the chance 
that both members of a given chromosome 
pair are free of lethals or semilethals is 
(0.75)- or 56%. 

What portion of individuals in the pop- 
ulation carry no lethal or semilethal on any 
member of autosomes II, III, and IV? This 
percentage is calculated as (0.75) 2 times 
(0.75)- times (0.75)-' or about 17%. 
However, if one considers the X and V 
chromosomes which can also carry such 
mutants, the frequency of lethal-semilethal- 
free individuals in nature is still lower. 
Moreover, when the subvital mutants (which 
comprise the most frequent mutant class de- 
tected) and the sterility mutants are also 
considered, it becomes clear that very few, 
if any, flies in natural populations are free 
of a detrimental mutant load. 

Genetic Loads in Man 

What is the genetic load in man? The vast 
majority of mutants are detrimental in homo- 
zygous condition (as already noted in Chap- 
ter 15). Since inbreeding increases the fre- 



quency of homozygosis, a comparison of the 
detriment produced in an inbreeding segment 
with that in a noninbreeding segment of a 
human population may provide us with an 
estimate of the genetic load present in het- 
erozygous condition. From the population 
records of a rural French population during 
the last century listing fetal deaths and all 
childhood and very early adult deaths we 
can compare the frequency of death to off- 
spring of unrelated parents with that of 
cousin marriages.-' The frequency of death 
to progeny from unrelated parents was .12, 
whereas it was .25 from cousin marriages. 
We are not concerned here with establishing 
the genetic or nongenetic cause of death in 
the normal outcrossed human population; 
however, it can be assumed that the extra 
mortality of .13 (.25 minus .12) has a ge- 
netic basis in the extra homozygosity result- 
ing from cousin marriage. This assumption 
is reasonable in the absence of any known 
nongenetic factor that tends to cause death 
to more or fewer offspring from marriages 
between cousins than from marriages be- 
tween unrelated parents. (These data would 
have a nongenetic bias if, for example, it 
were the custom — which it was not — that 
all children from cousin marriages are pur- 
posely starved.) 

Apparently, then, 13% more offspring 
died because their parents were cousins. 
The total amount of recessive lethal effect 
present in the population in heterozygous 
condition can be calculated as follows: recall 
(Chapter 15) that of all heterozygous genes, 
an extra l / 16 become homozygous in off- 
spring of cousin marriages. In the model 
half of the l / 16 , or y 32 , must have become 
homozygous for the normal genes and half 
of Y 1G , or y 32 , for their abnormal alleles. 
Therefore, to estimate the total heterozy- 
gous content of mutants which would have 
been lethal if homozygous, it is necessary to 

- Based upon an analysis of N. E. Morton, J. F. 
Crow, and H. J. Muller. 



218 



CHAPTER 16 



multiply .13 by 32. The resultant value of 
about 4 represents a 40095 chance that the 
ordinar) individual carried in heterozygous 
condition a genetic load of detrimental mu- 
tants which would have been lethal if homo- 
zygous. In other words, on the average, 
each person carried four lethal equivalents 
in heterozygous condition, or, lour times the 
number of detrimentals required to kill an 
individual if the genes involved somehow 
became homozygous. 

The preceding analysis did not reveal the 
number of genes involved in the production 
of the tour lethal equivalents. These lethal 
equivalents might have been due to the pres- 
ence in heterozygous condition of four re- 
cessive lethals, or eight mutants producing 
50% viability, or sixteen mutants with 25% 
viability, or any combination of detrimental 
mutants whose total was four lethal equiv- 
alents. Because of environmental improve- 
ments (better housing, nutrition, and med- 
ical care) since the last century, it is likely 
that the effect of the same mutants in present- 
day society would be expressed by some- 
what less than four lethal equivalents. For 
the same reason, the detrimental effects of 
these mutants in heterozygous condition are 
expected to be somewhat less at present than 
they were a century ago. For example, in 
the last century a particular hypothetical 
homozygous combination having variable 
penetrance and expressivity would have pro- 
duced no detectable effect 25% of the time; 
a detrimental effect — but not death before 
maturity — 15% of the time; and death be- 
fore maturity 60% of the time; today, the 
respective values would be 50%; 10%; 
40%. A century ago this combination 
would have produced .6 of a lethal equiva- 
lent; at present, the portion is .4. Notice 
also that the detriment not lethal before 
maturity would also have been reduced dur- 
ing this period from 15% to 10% or, speak- 
ing in terms of detrimental equivalents, what 
had been .15 would now be .10. Appar- 



ently the genes responsible for lethal equiva- 
lents and for detrimental equivalents must 
be the same, at least in part. 

It is also apparent that present-day man 
carries a genetic load. Some of those mu- 
tants transmitted to him arose in his parents 
(probably two of each five zygotes carry a 
newly arisen mutant, as mentioned on p. 
192), and others arose in his more remote 
ancestry. It has been calculated :i that, on 
the average, each of us is heterozygous for 
what is probably a minimum of about eight 
such mutant genes. This genetic load does 
not include the mutants carried in homozy- 
gous condition. What happens to this load 
of mutants in successive generations? 

Balanced vs. Mutational Loads 
To predict, in a general way, the fate of the 
"usual" mutant in the population, it is neces- 
sary to determine its "usual" phenotypic 
effect. 4 Since the typical mutant is detri- 
mental when homozygous — at least to some 
degree — the homozygous condition tends to 
eliminate it from the gene pool. But two 
opposite effects are possible for mutants 
when heterozyg3us (see Chapter 15): either 
the heterozygote is superior to both homo- 
zygotes (as is found for the sickling-causing 
gene in malarial countries), or the hetero- 
zygote is somewhat inferior to the nonmu- 
tant homozygote (as is true for most point 
recessive lethal heterozygotes). In the for- 
mer case the heterotic effect tends to in- 
crease the frequency of the mutant, and both 
the normal and mutant genetic variants are 
retained in the population gene pool at equi- 
librium. A population which normally re- 
tains more than one genetic (or chromo- 
somal) alternative in its gene pool, there- 
fore, exhibits balanced polymorphism in its 
phenotypes. This component of the genetic 
load is balanced, and is, therefore, a balanced 

■■ By H. J. Muller and by H. Slatis. 

•» See B. Wallace (1963), J. F. Crow and R. G. 

Temin (1964), and Th. Dobzhansky (1964). 



Genetic Loads and Their Population Effects 



219 



load. When the heterozygote is inferior to 
one homozygote, the heterozygous condition 
increases the rate at which the mutant is 
eliminated from the gene pool, and the pop- 
ulation shows unbalanced polymorphism and 
tends to become genotypically and pheno- 
typically monomorphic. This component of 
the genetic load, called the mutational load, 
is maintained in the population chiefly by 
recurrent mutation. Experimental evidence 
in Drosophila 5 and a statistical analysis of 
data for man ,; support the view that the 
great majority of point mutants are detri- 
mental when heterozygous. We shall, there- 
fore, consider most of the genetic load to 
be a mutational load. 

Genetic Death 

How is a mutant gene eliminated from the 
population? It need not be eliminated by 
the death of an individual, although some- 
times it is. A more general way to express 
the removal of a mutant gene from the gene 
pool is by genetic death — the failure of a 
mutant-carrying individual to produce de- 
scendants carrying the mutant. Thus, all an 
individual's genes, whether normal or mu- 
tant, suffer genetic death if that individual 
fails to produce children. Since mutants 
are stable, they are usually removed from 
the gene pool by genetic death and only 
occasionally by mutation. 

A person carrying a dominant lethal like 
retinoblastoma suffers genetic death (as well 
as physical death). In this case the mutant 
gene is eliminated from the population the 
generation in which it arises; it has, there- 
fore, only one generation of persistence. A 
dominant detrimental mutant with a selec- 
tion coefficient of .2 and, therefore, an adap- 
tive value of .8 as compared to normal, will 



n Based upon works of H. J. Muller and co-work- 
ers, C. Stern and co-workers, J. F. Crow and co- 
workers, I. H. Herskowitz and R. C. Baumiller, 
and others. 
11 Based upon an analysis by N. E. Morton. 



persist for five generations, on the average, 
before suffering genetic death; that is, given 
a population approximately the same in size 
for successive generations, in each genera- 
tion the mutant-containing individual has a 
20% chance of not transmitting the mutant. 
After this mutant arises, it sometimes fails 
to be transmitted the very first generation; 
it may suffer genetic death at the fifth gen- 
eration or at the tenth, but. on the average, 
the mutant persists five generations. The 
principle of persistence holds even though 
genetic drift, migration, or other factors 
cause fluctuations in the frequency of the 
mutant. 

Consider the fate in the population of a 
rare recessive lethal gene like the one pro- 
ducing juvenile amaurotic idiocy. Each 
time homozygosis for this gene occurs, it 
results in genetic death, and two mutant 
genes are eliminated from the gene pool. 
But consider the fate of heterozygotes which 
are 600 times more frequent (Chapter 15) 
and carry 300 times as many of these genes 
as do homozygotes. Since it is generally 
true that heterozygotes for a recessive lethal 
suffer genetic death about two per cent of 
the time (see p. 195), approximately .02 
times 600, or twelve, heterozygous people 
would suffer genetic death, thus involving 
the removal of 24 genes, twelve of them 
being the recessive lethal alleles. Accord- 
ingly, six times as many of these particular 
recessive lethal genes suffer genetic death 
in the heterozygote than in the mutant homo- 
zygote, even though the reduction in repro- 
ductive potential in the former type is only 
'.-o of that in the latter. 

It is apparent that the rarer a mutant is, 
the smaller will be the proportion of all 
genetic deaths it causes in homozygotes and 
the larger the proportion in heterozygotes. 
For rare mutants, then, natural selection re- 
moves mutant genes primarily via the ge- 
netic death of heterozygotes, the small 
amount of detriment being more important 



•220 



( IIAI'TER 16 



when heterozygous— from the population 01 

gene pool standpoint — than the greater det- 
rimental effect when homozygous. How- 
ever, each rare mutant, in terms of its etleet 
on reproductive potential, is equally harmful 
to a constant-sized population in that each 
eventually causes a genetic death. Thus, hy- 
poploid} 1 which aets as a dominant lethal per- 
sists onlv one generation before it causes 
a genetic death; a rare point mutant whose 
reproductive disadvantage is only U U) % will 
persist, on the average, one thousand gen- 
erations before causing genetic death. 

Consider, on one hand, the gross chromo- 
somal abnormality which kills in utero, de- 
stroying a life early. Neither the individual 
involved nor its parents suffer very long, 
since such deaths may occur as abortions 
which pass unnoticed. On the other hand, 
the heterozygous point mutants in individuals 
who are past the reproductive age — and, 
therefore, already have or have not suffered 
genetic death — will continue to subject these 
people to the previously and newly pro- 
duced, small phenotypic detriment of hetero- 
zygosity which adds to their aches, pains, 
and disease susceptibility. In this respect, 
then, the mutant with a small effect on re- 
productive potential can cause more suffer- 
ing than one with a large effect, for the 
longer the persistence, the more the damage 
in postreproductive life. In general, speak- 
ing not in terms of biological fitness but in 
terms of the total amount of suffering to 
which a human population is subject, point 
mutants with the smallest heterozygous 
detriment are the most harmful type of 
mutant. 

One might at first suppose that the 
amount of gene-caused human suffering can 
be reduced by medical science. This pos- 
sibility exists, particularly for an individual 
such as the diabetic who takes insulin; no 
doubt he is better off than he would be 
without medicine. But remember that this 
medicine does not cure the genetic defect. 



Moreover, by increasing the diabetic's repro- 
ductive potential, the medicine serves to 
increase the persistence of the mutants in- 
voked, and the genetic death which must 
eventually occur is only postponed to a later 
generation — each intervening generation re- 
quiring the same medication. The total 
amount of human sutlering would be re- 
duced only if medicine could correct the 
gene-produced defect. To correct all of the 
multiple effects of the mutant, the medicine 
would have to replace the primary product 
of the defective gene with normal product. 
Insofar as most, if not all, currently known 
medicines act later than this earliest stage 
in gene action (Chapter 6), they serve to 
alleviate only some detrimental effects, thus 
causing an increase in human suffering by 
increasing persistence. Unfortunately, this 
situation will continue until medical science 
is much further advanced. 

In view of the preceding discussion, we 
can assume that it is primarily the euploid 
or nearly euploid mutants which persist in 
the gene pool and are mainly responsible for 
changes in its composition during the course 
of evolution. By far the most common and 
most important class of such mutants is the 
point mutant. 

Mutation and Evolution 

In Chapter 1 5 we only suggested that muta- 
tions provide the raw materials for biolog- 
ical evolution. The reason for our hesitance 
in specifying evolution as the natural out- 
come of changes in gene pools was that the 
great majority of mutants, including point 
mutants, are harmful in homozygous or hemi- 
zygous condition. In this chapter and in 
Chapter 15, we indicated that most mutants 
are also detrimental when heterozygous. 
Under these circumstances, how can muta- 
tion provide the more adaptive genotypes 
postulated as necessary for evolution? It is 
true that for a given genotype under a given 
set of environmental conditions the great 



Genetic Loads and Their Population Effects 



221 



majority of point mutants are detrimental, 
and that, perhaps, only one point mutant in 
a thousand minutely increases the reproduc- 
tive potential of its carrier. Yet, provided 
the mutation rate is not too large and there 
is sufficient genetic recombination, these rare 
beneficial mutants offer the population the 
opportunity to become better adapted. 
Moreover, mutants which lower biological 
fitness under one set of environmental con- 
ditions may be more advantageous than the 
normal genes under different environmental 
circumstances. 7 For example, a mutant 
producing vestigial wings in Drosophila is 
clearly inferior to its normal genetic alterna- 
tive in an environment where flight ability 
is advantageous; but this mutant might be 
advantageous for Drosophila living on a 
small island where flight is not only un- 
necessary but harmful because insects that 
fly can be blown out to sea and lost. Con- 
sider a second example of this type. Several 
decades ago the environment was DDT-free. 
and mutants which confer immunity to DDT 
were undoubtedly less adaptive than the 
normal genetic alternatives present. But 
once DDT was introduced into the insect 
environment, such mutants — even if detri- 
mental in other respects — provided such a 
tremendous reproductive advantage over 
their alternatives that they became estab- 
lished in the population as the new wild- 
type genes. Still other examples can be 
cited involving antibiotic-resistant mutants 
in microorganisms, which in an antibiotic- 
free environment are less adaptive than the 
genes normally present. 

It becomes clear, then, that mutation pro- 
vides the opportunity for a population to 
become better adapted to its existing en- 
vironment. It also provides the raw mate- 
rials needed to extend the population's range 
to different environments, either those al- 
ready existing elsewhere or those that will 

•See Th. Dobzhansky (1964). 



arise through changes. A population that 
is already very well adapted to its present 
environment is appreciably harmed by the 
occurrence of mutation. But environments 
differ, and any given environment will even- 
tually change, so that a nonmutating popu- 
lation though successful at one time will, 
in the normal course of events, eventually 
face extinction. Mutation, therefore, is the 
price paid by a population for future adap- 
tiveness to the same or different environ- 
ments. We can now appreciate that muta- 
tion and selection, together with genetic 
drift and migration, are primarily responsible 
for the origin of more adaptive genotypes. 
We can also better appreciate the advantage 
of genetic recombination in speeding up the 
production of adaptive genotypes and the 
importance of the genetic mechanisms which 
regulate mutation frequency. 

Somatic Mutations 

In view of the preceding discussion, it is not 
at all difficult to predict the consequences of 
increasing the mutation frequency in human 
beings, an increase that doubtlessly is oc- 
curring as a result of our exposure to man- 
made penetrating radiations and certain re- 
active chemical substances. Man-made as 
well as spontaneous mutations can occur in 
either the somatic line or the germ line. 
Somatic mutants are, of course, restricted 
to the person in which they occur. The 
earlier the mutation occurs in a person's 
life, the larger will be the sector of somatic 
tissue to which the mutant cell gives rise. 

When an adult is exposed to an agent 
which causes a mutation to occur in a cer- 
tain percentage of all cells, the cells carrying 
induced mutants will usually be surrounded 
by nonmutant ones of the same tissue whose 
overall action produces a near-normal phe- 
notypic effect. When an embryo is exposed, 
a proportionally smaller number of its cells 
will mutate. Mutant embryonic cells can, 
however, give rise later to whole tissues or 



CHAPTER 16 



organs which arc defective; in such cases 
there is no compensator) action of normal 
tissue. Furthermore, since many mutants 
affect the rate of cell division, the earlier 
in development they occur, the more ab- 
normal the si/e o\' the resulting structure 
will be. It is understandable, then — assum- 
ing that cells at all stages are equally mutable 
— that the earlier somatic mutations occur 
in the development of an individual, the 
more damaging they will be to him. 

Newly arisen mutants produce almost all 
their somatic damage when heterozygous, 
since mutation involves loci which are usu- 
ally nonmutant in the other genome. Al- 
though somatic mutants cannot be trans- 
mitted to the next generation, they can lower 
the reproductive potential of their carriers, 
thus affecting the gene pool of the next 
generation. 

The damage which new mutants produce 
in a somatic cell depends upon whether or 
not the cell subsequently divides. Certain 
highly differentiated cells in the human body, 
like nerve cells or the cells of the inner lining 
of the small intestine, do not divide. In 
such cases, it is ordinarily difficult to detect 
mutations since the cells have no progeny 
classifiable as mutant or nonmutant. Non- 
dividing cells may be more or less mutable 
than those retaining the ability to divide. 
In any event, a variety of mutations can 
occur in nondividing cells, including point 
mutations which inactivate or change the 
type of allele present, as well as structural 
rearrangements of all sizes. Nevertheless, 
the nondividing cell remains euploid or 
nearly euploid, and the phenotypic detriment 
produced must be due almost entirely to 
point mutants in heterozygous condition and 
to shifts in gene position. Although this 
may considerably impair the functioning of 
nondividing cells and give the impression 
that they are aging prematurely, their sud- 
den and immediate death due to mutation 
is probably very rare. 



Although the same kinds of mutations 
occur in somatic cells that subsequently di- 
vide and in those that do not, nuclear divi- 
sion can result in gross aneuploidy (Chapters 
I I and 12). Accordingly, most of the phe- 
notypic damage of induced mutants in divid- 
ing cells is the result of aneuploidy — mostly 
the consequence of single breakages that fail 
to restitute. It should be noted that all 
known agents causing point mutation also 
break chromosomes. 

Germinal Mutations 

Since somatic cells comprise a population 
produced by asexual reproduction (cell divi- 
sion), the preceding discussion of the effects 
of somatic mutation is appropriate to this 
chapter. Consider next, in a general way, 
the consequences of increasing the frequency 
of mutations in the human germ line. The 
earlier that mutation occurs in the germ line, 
the greater the portion of all germ cells 
carrying the new mutant will be. Of course, 
the upper limit of gametes carrying a par- 
ticular induced mutant is usually fifty per 
cent. Consider the effect of exposing the 
gonads of each generation to an additional 
constant amount of high-energy radiation 
(Figure 16-3). The load of mutants pro- 
duced spontaneously is presumably at equi- 
librium — the rate of mutant origin equals 
the rate of mutant loss via genetic death. 
Beginning with the first generation to receive 
the additional radiation exposure, the mu- 
tant load increases with each generation 
until a new equilibrium is reached; at this 
point the higher number of genetic deaths 
per generation equals the higher number of 
new mutants produced each generation. If 
the additional radiation exposure ceases at 
some still later generation, the mutational 
load will decrease gradually (because of 
variations in persistence) via genetic deaths, 
until the old spontaneous equilibrium is 
reached again. 



Genetic Loads and Their Population Effects 



223 



GENETIC 
LOAD 




Spontaneous 



Radiation 
Exposure 



GENERATIONS 



figure 16-3. Genetic load and exposure to 
radiation. 



It is clearly important to learn in detail 
the genetic effects of high-energy radiation 
to which human populations are being sub- 
jected either purposely or circumstantially. 
In order to make the best evaluation, we will 
need to know much more about: the dis- 
tribution of the energy of various radiations 
in tissue; the exact amount of gonadal ex- 
posure to radiations of different types; the 
detriment of the induced mutants in hetero- 
and homozygous conditions; the persistence 
of mutants; and the different types and the 
frequencies of mutations that each kind of 
radiation produces in different stages of male 
and female gametogenesis. 

In the last respect, it is necessary to de- 
termine for various types of mutations, the 
relative mutagenicity of a concentrated dose 
and one given in a protracted manner. It 
is also necessary to learn as accurately as 
possible the mutability of spermatogonia and 
oocytes, because these are the stages in 
which the human germ cells producing the 
next generation remain for the longest pe- 
riod of time. It is suggested that the largest 
number of germ-line mutations occurs in 
oocytes. Because spermatogonia are con- 
stantly dividing, mutants producing a detri- 
mental effect may be selected against so that 
they are reduced in frequency by the time 
gametes are formed; the human female, how- 
ever, is born with all, or almost all, her 



future gametes already in the oocyte stage 
so that there is no parallel mitotic selection 
in this germ line. Not only do oocytes fail 
to undergo mitosis, but they remain rela- 
tively inactive for decades before becoming 
ova; as oocytes age during this period, they 
become disproportionately sensitive to spon- 
taneous mutation (at least to factors leading 
to aneusomy). 

Although at present we do not have as 
much information about any one of these 
factors as we would like, available informa- 
tion along these lines already gives us ap- 
proximate answers (see the references at the 
end of this chapter). It should be noted, 
therefore, that all values given in the dis- 
cussion below may be in error by as much 
as a factor of two or more. 

It has been a practice to discuss the germ- 
line effect of radiation in terms of the amount 
of increase any particular exposure would 
produce in our spontaneous mutation fre- 
quency. The general impression is held that, 
as a species, man is fairly well adapted to 
his spontaneous mutation rate, and that if 
this rate is doubled it will not threaten his 
survival. Accordingly, the question be- 
comes, how much man-made radiation would 
produce as many mutations as occur nor- 
mally? A United Nations report calculates 
that about 30 rads (roughly equal to 30 r) 
is sufficient to double the human sponta- 
neous mutation rate — the frequency per gen- 
eration. This amount is called the doubling 
dose. In a population of one million peo- 
ple, one rad delivered to the gonads, or sex 
organs, of each person is calculated to pro- 
duce between 100 and 4,000 mutants which 
could be transmitted to future generations. 
Thus, one rad of gonadal exposure for one 
generation will result in the birth of 100 to 
4,000 people with new heterozygous mu- 
tants. Affected descendants will occur for 
many generations, since only a small por- 
tion of the genetic deaths from these mutants 
will occur in the first generation. These 



224 



CHAPTER 16 



will not be evident when added to the 
number of genetic deaths resulting from 
spontaneous mutation. It' the one rad gon- 
adal exposure were repeated every genera- 
tion, an equilibrium would eventually be 
established in which, for every generation, 
100 to 4.000 people per million would show 
the effects of radiation-induced mutants in 
the form of genetie death. However, since 
the kinds of phenotypic effects produced by 
the radiation-induced mutants would be the 
same as those from mutants which occur 
normally, we would not be able to recognize 
specifically those people hurt by the radia- 
tion. 

What part of our normal load of mutants 
comes from naturally-occurring penetrating 
radiation? Since human beings receive 
about five rads in the course of a reproduc- 
tive generation — that is, in 30 years — it is 
possible that as little as % , or y 6 , of our 
mutations normally are radiation-induced. 

How much additional radiation are we 
exposed to during medical treatment? If 
medical use of radiation were to continue at 
its present level, it has been estimated that 
each person in the United States would re- 
ceive a total dose to the sex cells of about 
three r per generation. Of course, some 
people do not receive this amount of radia- 
tion, while others get considerably more. 
But this average radiation dose to the germ 
cells from medical uses alone is 60% of the 
amount received spontaneously and is rais- 
ing the mutation rate about 10% above the 
spontaneous rate. In the years to come, 
with increased use of radiation for diagnosis 
and therapy, the average dose from medical 
radiation might increase greatly. Already 
radioactive materials have been used in one 
million medical treatments in a single year. 
Many governments as well as private dental 
and medical groups are investigating such 
radiation exposure, and many ways of re- 
ducing radiation exposure without hinder- 
ing its usefulness are being instituted, 



It is difficult to determine the number of 
germ-line mutations resulting from the radia- 
tion associated with fallout following atomic 
explosions, because some radiation reach- 
ing the gonads could come from fallout on 
the ground, breathed in, or ingested with 
food. In the latter case, the distribution of 
particular radioactive substances in the body 
makes a large difference in the amount of 
radiation reaching the sex cells. With re- 
spect to sex cells, the three most important 
radioactive substances in fallout are cesium- 
137, strontium-90, and carbon- 14. Because 
cesium is distributed through the tissues — 
including the gonads — more or less evenly 
whereas strontium is preferentially localized 
in bone, we expect cesium- 137 to produce 
more gonadal radiation damage from in- 
gested fallout than strontium-90 produces. 

The period of time over which radioac- 
tive substances produce new mutations also 
varies. The induction of mutations by rela- 
tively short-lived radioactive substances, like 
strontium-90 and cesium- 137, is restricted 
almost entirely to a few generations. On 
the other hand, carbon- 14 — C-14 — is long- 
lived, having a half-life of 6,000 years. So, 
if the exposure to C-14 in the environment 
does not change, there will be about half as 
many new mutations induced after 200 gen- 
erations as there are in the first generation. 
Because of its abundance and long half-life, 
carbon- 14's potential for delivering radia- 
tion to the gonads has been calculated as 
being 4 to 17 times more than radioactive 
cesium's and strontium's combined and, 
therefore, carbon- 14 is capable of producing 
proportionally more point mutations. 

In the United States National Academy of 
Sciences — National Research Council report 
of 1956 (see References), the gonadal dose 
expected from fallout — if weapons of the 
same type continued to be tested at the same 
rate — was given as about 0.1 rad in the next 
thirty years. On the basis of the United 
Nations report, we could expect approxi- 



Genetic Loads and Their Population Effects 



225 



mately 10 to 400 mutations per million peo- 
ple. How much modification does this figure 
now need in order to bring it up to date? 
Before accurate estimates of germ-line mu- 
tational damage due to fallout can be ob- 
tained, many factors need be taken into ac- 
count, among them: 

1. Carbon- 14, whose long half-life was 
not considered in this report 

2. The changes in rate of testing (accord- 
ing to the United States Atomic En- 
ergy Commission, in 1958 alone the 
amount of fallout-producing radioac- 
tive material in the stratosphere was 
doubled by the numerous test explo- 
sions of nuclear weapons conducted by 
the United States and the U.S.S.R.) 

3. The unequal distribution of fallout in 
different parts of the world 

4. Reduction in decay taking place in the 
stratosphere since fallout is descending 
faster than expected 

5. Changes in the nature of bombs tested 
and in the location of the test sites 

6. The decrease in exposure as a result of 
the test ban treaty. 

Each month brings more of the data re- 
quired to estimate the fallout risk to the 
germ cells. Apparently, the possible dam- 
age has been underestimated. In 1953 the 
International Commission on Radiological 
Protection recommended — and various U.S. 
Government agencies adopted — 80 units as 
the maximum permissible concentration of 
strontium-90 in food. In 1958 the Com- 
mission recommended this maximum be 
lowered to 33 units, and the new value has 
subsequently been employed as a guideline 
by the U.S. Government. 

In principle, exposure to man-made radia- 
tion undoubtedly produces point mutants in 
the somatic and germ lines of man, but this 
possibility is not easy to demonstrate in prac- 
tice principally for two reasons: The first is 
that the expected point mutants are not 



qualitatively different from those which oc- 
cur spontaneously; the second is that the 
quantitative effect, although large in total, 
is small enough in any one generation to 
be masked by the general variability of hu- 
man genotypes and environment. Through 
the use of improved statistical methods, 
however, the evidence that radiation has pro- 
duced such genetic effects is becoming in- 
creasingly strong. On the other hand, clear 
proof that radiation can cause structural 
changes in human chromosomes does exist. 
With the recent perfection of cytological 
methods for studying human chromosomes 
and the evidence that aneusomy is a rela- 
tively frequent event in oocytes, it is likely 
that additional data will be forthcoming 
about the numbers and kinds of gross chro- 
mosomal mutations which different types 
and doses of radiation can induce in man. 

In discussing the genetic effects of low 
radiation doses, we recognized a danger 
which is not likely to be calamitous to the 
human gene pool; however, the very high 
radiation doses from a nuclear war could be 
disastrous, for if the whole body receives 
500 r in a short period of time, the chance 
is 50% that the affected person will die in 
a few months. If the person survives this 
period, his life expectancy is reduced by 
some years, probably because of somatic 
mutations, and children conceived after ex- 
posure will be handicapped by many detri- 
mental mutants. It is even possible, but 
not probable, that in a nuclear war enough 
radiation would be released to destroy the 
human species. 

Finally, it should be realized that we are 
being constantly exposed to man-made mu- 
tagenic chemical substances. Although it 
is very probable that we are getting fewer 
germ-line mutations from chemical sub- 
stances than from radiation, more somatic 
mutants may be produced by chemical sub- 
stances than by our present exposure to 
radiation. 



22(5 CHAPTER 16 

SUMMARY AND CONCLUSIONS 

Cross-fertilizing species carrj a large load oi mutants in heterozygous condition. The 

vasl majorit) ol them are detrimental when homozygous and to a lesser extent — when 
heterozygous, although some heterozygotes are superior to either homozygote. Other 

things being equal, almost all mutants are harmful to the same degree in that each 
eventually causes genetic death. Mutants producing the smallest detriment to repro- 
ductive potential cause the greatest total amount of suffering. More detriment and 
more genetic deaths occur in heterozygotes than in homozygotes for rare mutants. 
Persistence of a mutant in the population is inversely related to its selection coefficient. 

Mutation is the current price paid by a population for the possibility of having a 
greater reproductive potential in the same or a different environment in the future. 
So, despite the rarity of mutants which increase reproductive potential in a given en- 
vironment, mutation provides the raw materials for evolution. 

Natural and man-made penetrating radiations are undoubtedly causing mutations 
in our somatic and germ cells, increasing our load of detrimental mutants. This 
exposure, though harmful, is most likely no threat to man's survival as a species at 
present, although it might be in the future should the exposure become large enough. 
Further research is needed to accurately assess the effects of high-energy radiations 
and chemical substances upon man's mutation rate and well-being. 



REFERENCES 

Auerbach, C, Genetics in the Atomic Age, Fair Lawn, N.J.: Essential Books, 1956. 

Background Material for the Development of Radiation Protection Standards, Report 
No. 1, Federal Radiation Council, Washington, D.C.: U.S. Government Printing 
Office, 1960. 

Chu, E. H. Y., Giles, N. H., and Passano, K., "Types and Frequencies of Human Chro- 
mosome Aberrations Induced by X-rays," Proc. Nat. Acad. Sci., U.S., 47:830-839, 
1961. 

Crow. J. F., "ionizing Radiation and Evolution," Scient. Amer., 201:138-160, 1959. 

Crow, J. F., and Temin, R. G., "Evidence for Partial Dominance of Recessive Lethal 
Genes in Natural Populations of Drosophila," Amer. Nat., 98:21-33, 1964. 

Dobzhansky, Th., Evolution, Genetics, and Man, New York: John Wiley & Sons, 1955. 

Dobzhansky. Th., "How Do the Genetic Loads Affect the Fitness of Their Carriers in 
Drosophila Populations?" Amer. Nat., 98:151-166, 1964. 

Herskowitz, I. H.. "Birth Defects and Chromosome Changes," Nuclear Information, 
3 (No. 2): 1-2, 4, 1960. 

Krieger, H., and Freire-Maia, N., "Estimate of the Load of Mutations in Homogeneous 
Populations from Data on Mixed Samples," Genetics, 46:1565-1566, 1961. 

Morton, N. E.. 'The Mutational Load Due to Detrimental Genes in Man," Amer. J. 
Human Genet., 12:348-364, 1960. 

Muller, H. J., "Mutational Prophylaxis," Bull. N.Y. Acad. Med., 2nd Ser., 24:447-469, 

1948. 
Muller, H. L. "Radiation Damage to the Genetic Material," Amer. Scientist, 38:33-59, 

126, 399-425, 1950. 

Miintzing, A., "A Case of Preserved Heterozygosity in Rye in Spite of Long-Continued 
Inbreeding," Hereditas, 50:377-413, 1963. 



Genetic Loads and Their Population Effects 227 

Report of the United Nations Scientific Committee on the Effects of Atomic Radiation, 
New York: General Assembly Official Records: 13th Session, Suppl. 17 (A/3838), 
Chaps. 5-6, Annexes G-I, 1958. 

Selected Materials on Radiation Protection Criteria and Standards: Their Basis and Use, 
Joint Committee on Atomic Energy, Congress of the United States, Washington, 
D.C.: U.S. Government Printing Office, 1960. 

The Biological Effects of Atomic Radiation, Summary Reports, Washington, D.C.: 
National Academy of Sciences — National Research Council, 1956 and 1960. (See 
Reports of the Genetics Committee.) 

Wallace, B., "A Comparison of the Viability Effects of Chromosomes in Heterozygous 
and Homozygous Condition," Proc. Nat. Acad. Sci., U.S., 49:801-806, 1963. 

QUESTIONS FOR DISCUSSION 

16.1. Do you suppose that the mutations which occur in man serve a useful function? 
Why? 

16.2. Compare the fate of a mutational load in asexually reproducing populations that 
are haploid, diploid, and autotetraploid. 

16.3. Discuss the effect upon the gene pool of mutants restricted to the somatic line. 

16.4. Can the gene that comprises part of a detrimental equivalent also comprise part 
of a lethal equivalent? Explain. 

16.5. Give examples of balanced and unbalanced polymorphism in the genetics of man. 

16.6. What is the relation between phenotypic detriment, genetic death, and genetic 
persistence? 

16.7. Discuss the relative importance of point mutants and gross structural changes in 
chromosomes to the individual and to the population. 

16.8. What is the difference, in terms of mutation, between a maximum permissible 
dose and a doubling dose of ionizing radiation? Is any dose of any radiation 
safe from a mutational standpoint? Explain. 

16.9. Compare the genetic composition of the mutant load caused by fallout, by med- 
ical uses of radiation, and by atomic reactor accidents. 

16.10. Do you believe it is essential for the general public to become acquainted with 
the genetic effects of radiation? Why? 

16.11. What are some of the beneficial uses of radiation? Are any of these based upon 
the genetic effects of the radiation? If so, give one or more examples. 

16.12. One of the components of fallout is radioactive iodine, 1-131, which has a half- 
life of about a week. Discuss the genetic effects expected in the somatic and 
germ lines of persons exposed to fallout. 

16.13. Susceptibility to leprosy may be due to a single irregularly dominant gene. S. G. 
Spickett notes that leprosy is increasing in some human populations that have 
been free of it for many generations. List some factors which may be responsible 
for this finding. 

16.14. Is a genotype adaptive in man today, one which would have been adaptive 2,000 
or 20,000 years ago? Explain. 

16.15. "The danger of mutation lies primarily in the rate with which it occurs." Crit- 
icize this statement. 

16.16. How can you explain the finding that in the genus Drosophila apparently the 
heaviest genetic loads occur in common and in ecologically most versatile species, 
whereas the lightest loads are found in rare and in specialized species and in 
marginal colonies of common species? 



Chapter *17 

CHROMOSOMAL REARRANGEMENTS 
IN NATURE 

Oenothera ' 

The evening primrose, Oenothera (Figure 
17-1 ). is a common weed found along road- 
sides, railway embankments, and in aban- 
doned fields. It exists in nature in a number 
of pure breeding, self-fertilizing strains — 
each with a characteristic phenotype. These 
strains can be cross-fertilized in the labora- 
tory. If the two strains crossed are La- 
marckiana and biennis the outcome in F x is 
surprising. First, the F, are not all uniform 
in phenotype as one would expect from pre 
vious experience with crossbreeding two pure 
lines, but three distinct types (which we will 
call A, B, C). Second, upon self-fertiliza- 
tion each of these three F, types is thereafter 
pure breeding. If the Fi were hybrid, we 
would expect self-fertilization to produce 
recombinants and. therefore, more than one 
phenotype in its progeny. These two pe- 
culiarities are summarized in Figure 17-2 
where the results obtained from similar 
crosses with garden peas are shown side 
by side. 

We must conclude from the Oenothera 
results that self-fertilizing strains cannot au- 
tomatically be considered pure homozygous 
lines, despite any contrary impression gained 
in Chapter 1. In order to obtain three dif- 
ferent genotypes in F,. either Lamarckiana, 

1 Based upon work of H. DeVries. O. Renner. 
R. E. Cleland. F. Oehlkers, A. F. Blakeslee, J. 
Belling, S. Emerson, and A. H. Sturtevant. 

228 



or biennis or both must be heterozygous. 
Assume that Lamarckiana is heterozygous 
for a single pair of genes. If so, how can this 
strain produce only Lamarckiana upon self- 
fertilization'.' To do this would require that 
the heterozygote produces only heterozy- 
gote progeny. But suppose that self-fertili- 
zation does, as expected, produce the two 
homozygotcs, both of which are lethal. 
( Recall that for yellow mice — p. 69 — only 
one homozygote is lethal; the other is viable. 
In the present case the two different alleles 
would both have to act as recessive lethals. ) 
This hypothesis which predicts that one half 
of the zygotes die before becoming mature 
Lamarckiana, is supported by the finding 
that approximately one half of the ovules 
regularly fail to produce seed upon self- 




figure 17-1. Oenothera. (Courtesy of R. E. 
Cleland.) 



Chromosomal Rearrangements in Nature 



229 



fertilization — evidence that in nature La- 
marckiana is a permanent heterozygote in 
this respect, with a balanced lethal system. 
In this case both lethals kill the individual 
sometime before seed formation, in fact, the 
lethal kills at the time of fertilization or very 
soon thereafter, being in effect a zygotic 
lethal (Figure 17-3). 

Recall that some plants, including Oeno- 
thera, have a haploid gametophyte genera- 
tion. Permanent heterozygosis could be 
maintained also, if one allele were lethal to 
the male gametophyte and the other to the 
female (Figure 17-3). Consequently, game- 
tophytic lethals can also provide a balanced 
lethal system which prevents half of the 
ovules from producing seed. We have al- 
ready seen an example of this kind of lethal 
in the self-sterility gene in Nicotiana (p. 
60). In general, all strains of Oenothera 
found in nature, including biennis, have en- 
forced heterozygosity due to the zygotic and 
gametophytic lethals which produce bal- 
anced lethal systems. 

Does a balanced lethal system explain 
why the phenotype of Lamar ckiana, for ex- 
ample, is the only one produced in the prog- 
eny after self-fertilization? Since all sexual 
organisms so far studied have many pairs 
of genes, it would not seem reasonable that 
Oenothera has only a single pair of recessive 



PEAS 

Tall Dwarf 

I t 

Tall Dwarf 

I * 

Tall Dwarf 

\ / 

F Tall 

/ \ 

F 3 Tall; 1 Dwarf 



OENOTHERA 
Lamarckiana Biennis 

t I 

Lamarckiana Biennis 

♦ t 

Lamarckiana Biennis 
F A B C 

'III 

F A B C 



ZYGOTIC 
LETHAL 





# 






© 



figure 17-2. Comparative breeding results 
from garden peas and Oenothera. 



GAMETOPHYTIC 
LETHAL 



figure 17-3. Balanced lethal systems that 
enforce heterozygosity. 



lethal genes whose manifold (pleiotropic) 
effects produce the entire phenotype. It is 
more likely that many gene pairs exist which 
form a single linkage group, so that the dip- 
loid individual has one genome whose genes 
are all linked to one recessive lethal and 
another genome whose genes are all linked 
to the allelic lethal. 

In other words, Lamarckiana behaves as 
though it contains two complexes of linked 
genes. Within a strain these genes are com- 
pletely linked by some mechanism that pre- 
vents recombination, leaving the gametes 
with only two kinds of genotypes. The two 
gene complexes are so constant in natural 
populations of a strain that in the case of 
Lamarckiana they are given the names gau- 
dens and velans, and identified as gaudens. 
velans; the biennis strain as described by its 



230 



( II M'TER 17 



BIENNIS 








LAMARCKIANA 


Albicans IAI 








Gaudens IGI 


Rubens IRI 








Velans (VI 


A R 1 


@) 


GAMETES 


((3) 


) G V 



GAMETES ((o)) G 



GAMETES (O)) G 



/ \ 



FIGURE 17-4 {above). Balanced lethal gene 
complexes in O. biennis and O. Lamarckiana. 



figure 17-5 (below). Linkage groups in hy- 
brids from interracial crosses. 



m P B Sp Cu 



FLAVENS-CURVANS 

FLAVENS-PERCURVANS 

FLAVENS-FLECTENS 

FLAVENS-VELANS 

RUBENS-FLAVENS 

RUBENS-CURVANS 

CURVANS-VELANS 

RUBENS-VELANS 




gene complexes is albicans. rubens. Figure 

17-4 shows how these balanced lethal gene 
complexes are distributed generation after 
generation in biennis and Lamarckiana. All 
the recessive lethal alleles in either of the 
two strains cannot be identical to those in 
the other; if they were the F, from crossing 
them would consist of only two different 
phenotypes, whereas three types are actually 
obtained. We can conclude, therefore, that 
the balanced lethal system generally found 
in Oenothera involves either a multiple 
allelic series or several pairs of genes or both. 
Each of the three different Fi hybrids 
obtained from crossing Lamarckiana with 
biennis breeds true upon self-fertilization, 
showing that each hybrid contains two com- 
pletely linked gene complexes. This con- 
clusion may or may not be true, however, 
of the breeding behavior of other hybrids 
obtained from interracial crosses. This 
ambivalence is illustrated in Figure 1 7-5 
with the gene complexes present in the dif- 
ferent hybrids shown at the left. The dis- 
tribution of the various genetic markers (top 
of Figure 1 7-5 ) in the gametes of these 
hybrids was determined from breeding tests. 
For example, the curvans. velans hybrid pro- 
duced only two kinds of gametes though 
heterozygous for all these marker genes, the 
markers behaving as if they were all com- 
pletely linked. On the other hand, the 
Havens. velans hybrid produced four kinds 
of gametes. The genes R, m, and P (all still 
linked to each other) segregated independ- 
ently of the genes B and Sp (both still linked 
to each other), so that half of the gametes 
contained one of the two parental combina- 
tions, the other half carried one of the two 
recombinations. In this case, therefore, 
genes which belonged to a single linkage 
group in the parent races behaved as two 
linkage groups during the gametogenesis of 
their hybrid. Since 50% recombination oc- 
curred in gametogenesis of the hybrid, these 
results are not really explained by postulat- 



Chromosomal Rearrangements in Nature 



231 



ing that flavens (or velans) is actually a 
single linkage group which cannot undergo 
crossing over with its partner gene complex 
in the parent race, but which can do so when 
its partner is velans (or flavens). 

Tests of the hybrid containing flavens. 
curvans showed m and P still completely 
linked but segregating independently of 5, 
which was, in turn, segregating independ- 
ently of Sp and Cu, so that in this case three 
linkage groups existed. Perhaps more link- 
age groups would have been found with 
additional genetic markers. In all cases, 
however, a given hybrid combination always 
showed the same linkage groups in its game- 
togenesis. 

Because at least three linkage groups can 
be identified in certain interstrain hybrids 
(even though these act as one in the self- 
fertilizing parental strain), it is expected that 
the diploid Oenothera has at least three pairs 
of chromosomes and cytological examination 
confirms this genetic expectation — all of the 
Oenothera strains discussed in this chapter 
having seven pairs of chromosomes. (Oeno- 
thera gigas, the triploid mentioned on p. 
151, has 2 1 chromosomes. ) If the balanced 
lethal system is based upon a single pair of 
genes located on a single pair of homologs, 
this pair of chromosomes must be hetero- 
zygous in viable progeny. But this hetero- 
zygosity would not be expected for the other 
six pairs of chromosomes, if they segregated 
independently. Consequently, all gametes 
of O. biennis, for example, which carry the 
albicans complex recessive lethal should sim- 
ilarly be expected to carry the rubens or the 
albicans homolog in each of the other six 
cases of independent segregation. However, 
this distribution is not found. We could 
then suppose that each of the seven chro- 
mosome pairs is heterozygous for a different 
recessive lethal. Upon self-fertilization, 
such a genotype would produce only viable 
Fi like the parent. Since this explanation 
predicts that only about (U) 7 of all ovules 



should develop as seeds, it cannot be the 
correct one for Oenothera in which, as men- 
tioned, about 50% of all ovuies mature into 
seeds. 

A clue to the orderly segregation oi com- 
plete gene complexes in Oenothera may be 
found by cytological study of meiosis. The 
typical self-fertilizing Oenothera in nature 
does not form seven separate bivalents as 
expected, but, as seen clearly at metaphase 
I, forms a closed circle of 14 chromosomes 
synapsed end to end (Figure 17-6). At ana- 




figure 17-6. Circle of 14 chromosomes in 
Oenothera. Chromosome number is clear in 
upper cell where the circle has broken open. 
(Courtesy of R. E. Cleland.) 



232 



CHAP II R 17 




figure 17-7. Ralph E. Cleland points to zig-zag chromosomal arrangement 
at the start of anaphase I of an Oenothera having a circle of 14 chromosomes 
at metaphase I. {Photograph, 1959, courtesy of The Calvin Company.) 



figure 17-8. 
thera. 



Manner of chromosome segregation during meiosis of Oeno- 




Chromosomal Rearrangements in Nature 



233 



phase I, moreover, adjacent chromosomes 
in the circle go to opposite poles of the 
spindle, so that at the start of the separation 
the chromosomes assume a zigzag arrange- 
ment (Figure 17-7). If we assume that 
paternal and maternal chromosomes alter- 
nate in the circle, then all paternal chromo- 
somes would go to one pole and all ma- 
ternal chromosomes to the other. The com- 
plete linkage of all genes in a complex would 
be explained by such chromosome segrega- 
tion (if crossing over is rare), and the gam- 
etes produced by an individual would be 
identical to those which united to form it 
(Figure 17-8). 

If in an alternate segregation procedure 
maternal and paternal genomes separate, a 
circle should always contain an even number 
of chromosomes. Moreover, we could pre- 
dict that when one gene complex no longer 
behaves as a single linkage group, it will 
also no longer form a single circle of four- 
teen chromosomes with the other gene com- 
plex. Fourteen chromosomes can be ar- 
ranged fifteen different ways in circles (com- 
posed of even numbers of chromosomes) 
and pairs as shown in Figure 17-9. Indeed, 
when various race hybrids are made, all fif- 
teen types and no others are found at meta- 
phase I — any particular hybrid always form- 
ing the same meiotic configuration. (The 
top cell in Figure 17-7 shows an inner 
circle of four and an outer circle of ten 
chromosomes.) If what has been supposed 
about alternate segregation is true, it should 
also follow that even though alternate chro- 
mosomes within a circle show complete link- 
age with each other, such linkage groups 
should segregate independently of other link- 
age groups consisting of chromosomes either 
in separate circles or in separate pairs. This 
expectation can be tested by comparing the 
number of genetically determined linkage 
groups in the different hybrids of Figure 
17-5 with the chromosome arrangements 



Q 14 Q 10, 2 Po.rs 

O 10 ' O 4 O 6 - O 4 - 2Pairi 

(7) 8, (JJ) 6 Q 8, 3 Pairs 

O •• O <• O 4 O 4 - O 4 - 3 *- 

Q\ 12, 1 Pair Q 6, 4 Pairs 

8, Q4, 1 Pair O 4 ' 5 Pairs 

6, (~) 6, 1 Pair 7 Pairs 

O "' O 4 ' O 4 ' } Pair Q=CIRCIE 

figure 17-9. Circle and pair arrangements 
possible for Oenothera chromosomes. 



seen cytologically during their meiosis. Such 
a comparison reveals that the number of 
separate groups of chromosomes observed 
in meiosis is always equal to, or greater than, 
the number of linkage groups detected ge- 
netically. In fact, whenever a sufficient 
number of genetic markers are used, the 
number of linkage groups always equals the 
number of chromosome groups. 

Although the preceding discussion indi- 
cates that a rather unique segregation of 
alternate chromosomes in a circle and the 
presence of balanced lethal systems can ex- 
plain most of the unusual genetic behavior 
of Oenothera, other matters still need ex- 
planation. What causes these chromosomes 
to form circles in the first place? A clue to 
this, contained in the observation made on 
p. 172 (see also Figure 12-6), is that two 
pairs of nonhomologs will be associated as 
a double tetrad during synapsis if a recip- 
rocal translocation involving them is present 
in heterozygous condition. Figure 17-10 
illustrates this situation in Oenothera. All 
Oenothera chromosomes are small, are 
roughly the same size, and have median 
centromeres. To help us identify homol- 



234 



CHAPTER 17 



1 2 



13 3 4 



r? 




3 3 




I 



5 5 



FIGURE 17-10. Heterozygous reciprocal trans- 
locations and circle formation. Chromatids 

arc not shown. 



OgOUS chromosomes, the ends of dilTerent 
chromosomes in a genome are given dif- 
ferent numbers. Suppose, at some time in 
the past, a dicentric reciprocal translocation 
occurred between the tips marked 2 and 3 
(Figure 17-10A, B). This rearrangement 
in heterozygous condition ( B ) would pro- 
duce an X-shapcd configuration at the time 
of synapsis in prophase I (C) and a circular 
appearance at metaphase I-early anaphase 
1 (D). In this way a circle of four chro- 
mosomes would be produced. 

If a second reciprocal translocation oc- 
curs between any chromosome arm in a cir- 
cle of four and an arm of some other pair 
of chromosomes, a circle of six chromosomes 
will form in the individual heterozygous for 
both reciprocal translocations. This type of 
formation is illustrated in Figure 17-10D, 
E; D shows the configuration before arms 
4 and 5 have exchanged, E shows the circle 
of six produced in meiosis after this ex- 
change. Still larger circles can be formed 
by successive interchanges of this type; six 
such interchanges are required to form a 
circle of 14 chromosomes. The presence 
of reciprocal translocations in heterozygous 
condition could explain how various sized 
circles containing even numbers of chromo- 
somes are produced in Oenothera. 

Although the cytogenetic analysis of Oeno- 
thera is known in some detail, the picture 
is, however, not yet complete. One of the 
questions remaining is: What is the mech- 
anism whereby alternate chromosomes in a 
circle proceed to the same pole during meio- 
sis? No fully acceptable answer to this ques- 
tion has yet been given. A second question 
stems from the fact that almost all the dilTer- 
ent strains or races of Oenothera found in 
nature form a circle of 14. Are the six 
translocations involved the same in all races? 
No — for if they were, viable hybrids be- 
tween races would form either circles of 14 
or seven separate chromosome pairs at meio- 



Chromosomal Rearrangements in Nature 



235 



sis. That all the configurations in Figure 
17-9 are found in meiosis of such hybrids 
must mean that different gene complexes 
differ from each other in the specific ways 
that their chromosome arms have become 
translocated. Many thousands of ways are 
possible for 14 ends to be arranged in seven 
groups of two. How can we determine the 
number of these different arrangements oc- 
curring in nature? 

We can start by choosing a particular gene 
complex — calling it the "standard'" — and 
considering its chromosome ends to be 1-2, 
3_4, 5_6, 7-8, 9-10, 11-12, and 13-14. 
Normally, that is, in nature, this complex 
would form a circle of 14 with the other 
gene complex, which would therefore have 
no chromosome with the same pair of num- 
bered ends as any chromosome in the stand- 
ard complex. Proceeding further, we form 
a series of interracial hybrids with the stand- 
ard as one of the complexes and score the 
meiotic chromosome arrangements of the 
hybrids. Suppose in one case that the hy- 
brid forms five pairs and a circle of four. 
This result must mean that the ends of 5 
chromosomes are in the same order in the 
complex under test as in the standard, but 
that they are in a different order in the re- 
maining two chromosomes. Although there 
was previously no reason to assign ends 1-2 
and 3-4 of the standard complex to any par- 
ticular chromosomes, we can presently assign 
these ends arbitrarily to the two standard 
chromosomes in the circle of four. The 
chromosomes in the circle from the complex 
under test can then be called 2-3 and 4-1 
(or 2-4, 1-3). In this way the composition 
of ends of two chromosome pairs is specified 
permanently. The top of Figure 17-11 
shows the standard and tested complexes (of 
our example) synapsed according to iden- 
tical numbers. 

Call the complex just tested A. Suppose 
another complex, B, is made hybrid both 



A. COMPLEXES DIFFERING BY ONE RECIPROCAL TRANSLOCATION 

1.2 3.4 5.6 7.8 9.10 11.12 13.14 

\ / \ II II I I I I I I 

23 41 56 7.8 9.10 11 12 13.14 



B COMPLEXES DIFFERING BY SIX RECIPROCAL TRANSLOCATIONS 
1.2 3.4 5.6 7.8 9.10 11.12 13.14 

2.3 4.5 6.7 8.9 10.11 12.13 14.1 



C. MURICATA RACE'S ACTUAL COMPOSITION OF Q 14 
1.2 3.4 6.5 13.12 7.11 10.9 8.14 

\ / \ / \ / I / I / \ / 1 
2.3 4.6 5.13 12.7 11.10 9.8 14.1 



A and B are theoretical 



figure 17-11. Arrangement of chromosome 
ends in different Oenothera complexes. 



with the standard and with A. The meiotic 
configuration of the hybrid may specify 
other ends of A, B, and the standard com- 
plexes. Such procedures can be carried out 
until all of the standard's chromosomes are 
specified and the complete order of all four- 
teen ends determined for any other complex. 
In this manner, we can verify that a circle 
of fourteen is produced in many different 
ways in nature; a hypothetical and an actual 
example is shown in the central and lower 
parts of Figure 17-11. In fact, of 350 com- 
plexes analyzed, more than 160 different 
segmental arrangements have been found. 
All these results are consistent with the hy- 
pothesis that during the course of evolution, 
the ends of Oenothera chromosomes have 
been shuffled many times in different ways 
by reciprocal translocation. A most convinc- 
ing test of the reciprocal translocation inter- 
pretation would be the ability to predict the 
meiotic chromosomal arrangement to be 
found in a hybrid not yet formed. This 
type of predicting has been done many times 
and all such expectations have been verified. 
At various points in this discussion Oeno- 
thera's behavior has seemed exceptional, ap- 



236 



( II MTI.R 17 



parently violating our concepts of pure lines 
and independent segregation. More com- 
plete analysis has shown, however, that 
( Oenothera's failure to behave as expected 
was due to the operation of other, already 
known, genetic events. Oenothera is an ex- 
ception which should be treasured; for in 
the exact correspondence between its atypical 
genetics and its atypical cytology, it is an 
outstanding example of the validity of the 
chromosome theory of transmission genetics. 

Three aspects of the cytogenetic behavior 
of Oenothera are disadvantageous under 
many circumstances: reciprocal transloca- 
tions; recessive lethals; and self-fertilization. 
By combining all three of these disadvan- 
tages in one plant, however, Oenothera's sur- 
vival value is probably greater than it would 
be without them. The self-fertilization 
mechanism involves bringing the stigma 
down to the level of the anther, so that a 
much heavier pollination is attained than 
would be likely were the plant pollinated by 
insects. This self-fertilization mechanism 
offsets the 50% mortality due to balanced 
lethals. These lethals, together with the re- 
ciprocal translocations and alternate segre- 
gation, prevent the homozygosity usually 
consequent to self-fertilization, enforce het- 
erozygosity, and produce maximum hybrid 
vigor. 

The great survival value of Oenothera is 
demonstrated by the distribution of this 
genus: it can be found from the southern 
tip of South America to the far reaches of 
Northern Canada and from the Atlantic 
Ocean to the Pacific. It is interesting to 
note that the most numerous sections of the 
genus and those which have ranged the 
farthest are the ones with large circles, bal- 
anced lethals and self-pollination. 

Drosophila 

Although reciprocal translocations have 
played an important role in the evolution 
of Oenothera, it might be claimed that this 



X >x 



>,< >,< >r 




'(* T T 




V s 


>> ^i 4 



figure 17-12. Chromosome configurations in 
several Drosophila species. 



genus is an unrepresentative example of the 
importance of chromosomal rearrangements 
in evolution because its cytogenetic behavior 
is so unorthodox. Hundreds of different 
species of Drosophila occur in nature. These 
species can be compared ecologically, mor- 
phologically, physiologically, and biochem- 
ically. For those species able to interbreed, 
recombinational genetic properties can also 
be compared; banding patterns of salivary 
gland chromosomes and the appearance of 
chromosomes at metaphase of different spe- 
cies are additional areas of comparison. 
After all available information of this kind 
is gathered, it is possible to arrange the chro- 
mosomes of various species on a chart so 
that those closest together are more nearly 
related in descent — evolution — than are those 
farther apart.- This arrangement is illus- 
trated in Figure 17-12 which shows the 
karyotype — the haploid set of chromosomes 

- Based upon work of C. W. Metz and others. 



Chromosomal Rearrangements in Nature 



237 



at metaphase — including the X but not the 
Y chromosome for different Drosophila spe- 
cies or groups of species. The karyotype 
of the melanogaster species group, for ex- 
ample, is shown in row 2, column 1; the 
bottom chromosome is the rod-shaped X, 
the two V's are the two large autosomes 
(II and III), and the dot represents the tiny 
chromosome IV. In the other karyotypes, 
whole chromosomes or chromosome arms 
judged to be homologous are placed in the 
same relative positions. What can be learned 
from a comparison of these karyotypes? 

Since the amount of detail in a metaphase 
chromosome is limited basically to size and 
shape, one cannot expect to discern any 
small-sized rearrangements at this stage. 
Accordingly, regardless of their importance, 
small rearrangements involving duplication, 
deficiency, shift, transposition, inversion, 
and translocation cannot be detected on the 
chart. Even a large paracentric inversion 
is undetected at metaphase, since it does not 
change the shape of the chromosome. Other 
gross structural changes, however, can be 
detected. In row 4 the chromosome pat- 
terns in columns 2 and 3 seem identical, 
except that a pericentric inversion has 
changed a rod to a V, or vice versa. (Peri- 
centric inversions always change the rela- 
tive lengths of the arms when the two breaks 
are different distances from the centromere.) 
Compare the karyotype for melanogaster 
(row 2, column 1 ) with the one to its right 
(row 2, column 2). A V-shaped autosome 
in melanogaster appears as two rods in its 
evolutionary relative. (Note also that the 
dot chromosome is missing.) In the next 
karyotype to the right (row 2, column 3), 
two rods have combined to form a V that 
is different from either of the two V's in 
melanogaster. 

Other examples in this chart indicate that 
two rod-shaped chromosomes have formed 
a V-shaped chromosome, or a V has formed 
two rods. Consider first how a V can origi- 



nate from two rods (Figure 17-13). Re- 
call that a rod-shaped chromosome typically 
has two arms, though one is very short. The 
short arm may not be noticeable at meta- 
phase or anaphase; however, its presence 
may be demonstrated either cytologically at 
an earlier or later stage of the nuclear cycle, 
or genetically by studying genetic recom- 
bination. Suppose two rods are broken near 
their centromeres, one in the long arm of 
one chromosome, the other in the short arm 
of the other chromosome. If the long acen- 
tric arm of the first chromosome becomes 
joined to the long centric piece of the sec- 
ond, a V is formed. Notice that this union 
involves the joining of two whole or almost- 
whole arms in a eucentric half-translocation. 
The remaining pieces may join together to 
form a short eucentric chromosome, thereby 
completing a reciprocal translocation; or 
they may not join. In either instance, if the 
short pieces are lost in a subsequent nuclear 



/* 



\ 



HALF (OR RECIPROCAL) TRANSLOCATION 



/\ 



figure 17-13. Formation of a V-shaped chro- 
mosome from two rod-shaped chromosomes. 






RECIPROCAL TRANSLOCATION 




PARACENTRIC DELETIONS 




figure 17-14. Formation of two rod-shaped 
chromosomes from a V-shaped chromosome 
and a Y chromosome. 



division and the number of genes lost is 
small enough, the absence of these parts 
may be tolerated physiologically by the or- 
ganism. 

The reverse process, the formation of two 
rods from a V, necessitates the contribu- 
tion of a centromere from some other chro- 
mosome. In Drosophila, this second chro- 
mosome may be the Y (Figure 17-14). 
Suppose the V is broken near its centromere 
and the Y is broken anywhere. Should a 
eucentric reciprocal translocation follow, two 
chromosomes would be produced, each hav- 
ing one arm derived predominantly from the 
Y. If subsequent paracentric deletions oc- 
cur in these Y-containing arms, rod shapes 
will result, thereby completing the change 
from a V to two rods. Note that almost 
every part but the centromere of the Y chro- 



CHAPTER I 7 

mosome is eventually lost in this process. 
But this loss may have little or do disadvan- 
tage to the Drosophila. since the Y carries 
relatively few loci and is primarily concerned 
with sperm motility. For example, this 
series of mutations may be initiated in the 
male germ line, producing two chromosomes 
— each containing part of the Y. Deletion 
of Y parts can occur without detriment if 
these chromosomes happen to enter the fe- 
male germ line; they may stay in the male 
germ line provided that a regular Y chro- 
mosome is included in the genotype in due 
time. The small IV chromosome in melano- 
gaster, whose monosomy is tolerated in either 
sex, may also contribute a centromere in 
the process of changing a V to two rods by 
an identical or similar series of mutational 
events. 

Karyotype comparisons of Drosophila 
confirm the expectation (Chapter 12), that 
whole arm translocations are able to survive 
in natural populations. Such rearrange- 
ments and pericentric inversions are ex- 
tremely useful in helping us establish evolu- 
tionary relationships among different species. 
But it should be emphasized that this kind 
of information by itself does not prove that: 

1 . The formation of different species is a 
primary consequence of the occurrence 
of these rearrangements 

2. These rearrangements are of secondary 
importance in species formation 

3. These mutational events occur after 
species formation is complete. 

As exemplified by Oenothera and Dro- 
sophila, we have seen that gross chromo- 
somal rearrangements of various types have 
persisted in the evolutionary course of differ- 
ent groups of organisms. For this reason 
it would perhaps be wise at this point to 
retrain from predicting — except generally as 
in Chapter 12 — which, if any, structural 
changes might be associated with the evolu- 
tion of other particular groups of organisms. 



Chromosomal Rearrangements in Nature 



239 



SUMMARY AND CONCLUSIONS 

Although the cytogenetics of Oenothera has several unusual aspects, present knowledge 
renders these differences quite understandable; consequently, Oenothera provides an 
outstanding confirmation o\ the validity of the chromosomal basis for genetic material. 
Evolution in this genus is intimately associated with self-fertilization, balanced lethals, 
and numerous reciprocal translocations. 

Pericentric inversions (which change chromosome shape) and whole arm reciprocal 
translocations (which lead to changes in chromosome number) have been frequent in 
the past evolutionary history of Drosophila. 

REFERENCES 

Cleland, R. E., "A Case History of Evolution," Proc. Indiana Acad. Sci. (1959), 69: 
51-64, 1960. 

Cleland, R. E., "The Cytogenetics of Oenothera," Adv. in Genet., 11:147-237, 1962. 

Patterson, J. T., and Stone, W. S., Evolution in the Genus Drosophila, New York: Mac- 
millan. 1952. 

White. M. J. D., "Cytogenetics of the Grasshopper Moraba scurra, VIII." Chromosoma, 
14:140-145, 1963. 



Hugo De Vries (1848-1935), pio- 
neer in the study of mutation and 
Oenothera genetics. (From Genet- 
ics, vol. 4, p. 1, 1919.) 




240 CHAPTER 17 

QUESTIONS FOR DISCUSSION 

17.1. Whal evidence can you present for saying that the genes which make up the 
balanced lethal system in Lamarckiana arc different from those in biennisl 

17.2. Discuss the following statement: "All evening primroses found in nature are 
constant hyhrids." 

17.3. With respect to chromosomes, how does the origin of a circle differ from the 
origin of a ring? 

17.4. Can a circle contain an odd number o\ chromosomes? Explain. 

17.5. What new investigations regarding the genetics and or cytology of Oenothera 
has this chapter suggested to you? 

17.6. List the genetic principles you could have deduced had Oenothera been the only 
organism studied so far. 

17.7. If this chapter contains no new principles of genetics, why do you suppose it 
was written? 

17.8. Curly-winged Drosophila mated together always produce some non-curly off- 
spring. Plum eye-colored flies mated together always produce some non-plum 
offspring. But, when flies that are both curly and plum are mated together, only 
flies of this type occur among the offspring. Explain all three kinds of results 
and define your symbols. 

17.9. (a) Draw a diagram representing a heterozygous whole-arm translocation in 
Drosophila at the time of synapsis. Number all chromosome arms involved, 
(b) What would be required for a mating between two flies with this constitu- 
tion to produce offspring flies only of this type? 

17.10. Do you suppose that the preservation of heterozygosity has an adaptive ad- 
vantage in Oenothera? In other organisms? 

17.1 1. Discuss the evolutionary flexibility of the genus Oenothera and Drosophila. 

17.12. Is the balanced lethal system in Oenothera part of its genetic load? Explain. 
If so, are the lethals components of a balanced load or a mutational load? 
Explain. 

17.13. Compare the genetic effects of ionizing radiation on populations of Oenothera 
and Drosophila. 

17.14. Explain how a Drosophila zygote formed with a sperm carrying a centric, 
grossly-deleted Y chromosome can develop into a fertile male. 



Chapter *18 

RACES AND THE ORIGIN 
OF SPECIES 



I 



'n cross-fertilizing species, differ- 
ent individuals in a population 
are heterozygous for different 
genes (see Chapter 16), even though the 
gene pool is at equilibrium with the factors 
that cause shifts in gene frequency — namely, 
mutation, selection, drift, and migration. In 
other words, in reaching genetic equilibrium, 
all the members of cross-fertilizing popula- 
tions do not eventually become homozygotes, 
nor do they all become heterozygotes. Such 
populations, therefore, do not become either 
genetically pure or uniform with the passage 
of time. 

Although any given population is poly- 
morphic for some genes, it is not necessarily 
polymorphic with regard to a particular gene. 
For example, Indians in South America are 
almost all of O blood type, being homozy- 
gous (//) in this respect, but have a poly- 
morphic pool with respect to other genes. 
Moreover, an allele, like I B , may be rare 
or absent in one population, as in certain 
North American Indians, and relatively fre- 
quent in the gene pool of another population, 
as in central Asia. Thus, populations lo- 
cated in different parts of the world may 
differ both in the types and frequencies of 
alleles carried in their gene pools. For many 
purposes it is desirable to identify a popula- 
tion with certain gene pool characteristics as 
a race. 

Races 

An investigator may choose to define races 
only according to the distribution of the / /; 
241 



gene for ABO blood type. He might define 
populations that do or do not contain I' : 
in their gene pool as different races. On 
this basis there would be only two races of 
man, the South American Indian (without 
I 11 ) and all the other people (with l B in their 
gene pool). 

On the other hand, an investigator may 
decide to define races on the basis of the 
relative frequency of i and / /; in the popula- 
tion. The frequency of these alleles in the 
gene pool has been determined for many 
populations all over the world. The results 
show that in western Europe, Iceland, Ire- 
land, and parts of Spain, three-fourths of the 
gene pool is i, but this frequency begins to 
decrease as one proceeds eastwardly from 
these regions. On the other hand, I B is most 
frequent in central Asia and some popula- 
tions of India but becomes gradually less and 
less frequent as one gets farther away from 
this center. Since the change in frequency 
of these alleles is gradual, any attempt to 
sharply separate people into races having 
different gene frequencies would be arbitrary. 

In practice, therefore, the number of races 
recognized is a matter of convenience. For 
some purposes separating mankind into only 
two races is adequate; for other reasons, as 
many as two hundred have been recognized. 
As a rule, most anthropologists recognize 
about half a dozen basic races but may in- 
crease the number to about thirty when con- 
sidering finer population details. Regardless 
of the number of races defined, however, 
each is best characterized according to the 
genes it contains. Since the people in a 
population are either A, B, AB, or O in 
blood type and intermediates do not occur, 
no average genotype exists for the ABO 
blood group, nor is there an average geno- 
type for any other polymorphic gene. Be- 
cause a population has no average genotype. 
a race should be defined according to the 
relative frequency of alleles contained in its 
gene pool. Without an average genotype, 



242 



CHAPTER 18 



a race cannot have an average phenotype; 
accordingly, it is futile to try to picture a 
typical (average) member of any race. 

Other blood traits, besides ABO blood 
group, whose genetic basis is understood. 
are also useful in characterizing races. In 
fact, it is valid to utilize any phenotypic 
difference due to a genetic difference. For 
example, in delimiting races one can employ 
certain genetic differences in color of hair. 
eyes, and skin, and differences in stature and 
head shape; one should avoid using pheno- 
ls pic ditTerences whose genetic basis is un- 
proved, for the environment itself can cause 
phenotypic differences (Chapter I ). Re- 
member also that the same phenotypic result 
may be produced by different genotypes be- 
cause of gene interaction in dominance and 
epistasis (Chapter 4). 

Knowledge of the distribution of genes 
for ABO blood types in different populations 
provides important information to geneti- 
cists, anthropologists, and other scientists. 
To what can the different distributions be 
attributed? Since people do not choose their 
marriage partners on the basis of their ABO 
blood type, and since there does not seem 
to be any pleiotropic effect making persons 
of one blood type sexually more attractive 
than those of another, it is very likely that 
mating is at random with respect to ABO 
genotype. However, in other respects some 
evidence indicates that different ABO geno- 
types do not have the same biological fitness. 
Differential mutation frequencies can also 
explain part of the differences in gene dis- 
tribution. During the past few thousand 
years the greatest shift in ABO gene fre- 
quencies of different populations has prob- 
ably been the result of genetic drift and 
migration. In fact, the paths of past migra- 
tions can be traced by utilizing — along with 
other information — the gradual changes in 
the frequencies of ABO and other blood 
group genes in neighboring populations. 



It has already been mentioned (p. 209) 
that different paracentric inversions are 
found in natural populations of D. pseudo- 
obscura. All oi these flies are very similar 
phenot\ pically, even though their chromo- 
somal arrangements are different. Sample 
populations of this fly in the southwestern 
part of the United States (Figure 1S-1) 
have been studied to determine the relative 
frequency of these inversions." California 
populations proved to be rich in the inver- 
sion types called Standard and Arrowhead. 
Eastward, in nearby Arizona and New Mex- 
ico, the populations contain relatively few 
Standard and Pikes Peak chromosomes, 
most chromosomes having the Arrowhead 
arrangement. Finally, in still more easterly 
Texas, one finds almost no Standard and 
some Arrowhead with most chromosomes 
being of the Pikes Peak type. 

The shift in the frequency and type of 
inversions in the three different geographic 
regions cannot be explained as the result of 
differential mutation, since the spontaneous 
mutation rate for inversions is extremely 
low. Moreover, since there is no indication 
that the gene flow among these populations 
has changed appreciably in the recent past, 
migration rates have probably had a rela- 
tively small influence upon genotypic fre- 
quencies; there is also no indication that 
genetic drift has had a major role in causing 
the differences in inversion frequency in the 
three areas. These observations lead us to 
suppose that the primary basis for these pop- 
ulation differences lies in the different adap- 
tive values which different inversion types 
confer on individuals in different territories. 
Despite the absence of any obvious morpho- 
logical effects, these inversions prove to have 
different physiological effects in laboratory 
tests; different inversion types survive best 
in different experimental environments. Since 

1 Bused upon work of Th. Dobzhansky and col- 
laborators. 



Races and the Origin of Species 



243 




figure 18—1. Distribution of inversion types 
in D. pseudoobscura collected in the South- 
western United States. (After Th. Dohzhan- 
sky and C. Epling.) 



these inversion types show different adaptive 
values in the laboratory, it is reasonably cer- 
tain that they do so in nature too. Accord- 
ingly, natural selection is primarily responsi- 
ble for the inversion differences among the 
three geographic populations, which can be 
defined as three different races. 

Similar results have been obtained with 
three California races of the cinquefoil plant 
species, Potentilla glandulosa, which live at 
sea level, mid-elevation, and the alpine zone. 
The sea level race is killed when grown in 
the alpine environment, whereas the alpine 
race grown at lower elevations proves less 
resistant to rust fungi than the lower-eleva- 
tion races. Such experiments show that dif- 
ferent races are adapted to their own habitats 
but not to others. The inorganic and organic 
environment — including its organisms — is 



different in different parts of the territory 
occupied by a species. Clearly, then, no 
single genotype will be equally well adapted 
to all the different environments encountered 
within a particular territory. One way in 
which a cross-fertilizing species can attain 
maximal biological fitness as a whole is to 
remain genetically polymorphic and sepa- 
rate into geographical populations or races 
which differ genetically. 

Whenever, as in all of the examples dis- 
cussed so far, different races of a cross- 
fertilizing species occupy geographically sep- 
arate territories, they are said to be allo- 
patric; different races occupying the same 
territory are said to be sympatric. In the 
absence of geographical separation, what 
factors operate to keep sympatric races from 
hybridizing to become one race? One may 
find the answer by considering the fate of 
races — originally allcpatric — which have be- 
come sympatric, a kind of change which has 
occurred in man. Several thousand years 
ago, mankind was differentiated into a num- 
ber of allopatric races. With the develop- 
ment of civilization and improved methods 
of travel, many of these races have become 
sympatric. Gene exchange in the now- 
sympatric races, however, is sometimes in- 
hibited by social and economic forces, so 
that some of these races continue to main- 
tain their identity. Domesticated plants and 
animals provide another example of what 
can happen when allopatric races become 
sympatric. Many different breeds, or races, 
of dogs originally allopatric are now found 
living in the same locality. Yet these now- 
sympatric races do not exchange genes with 
sufficient frequency to form a single mongrel 
breed, or race, because their reproduction 
is controlled by man. It should be realized 
that, under other circumstances, such allo- 
patric races which become sympatric can 
form a single polymorphic race via cross- 
breeding. 



244 



CHAPTER 18 



Speciation Involving One Species 

A species of cross-fertilizing organisms usu- 
ally consists of a Dumber of races adapted 

to the different environments of the terri- 
tories thej occupy. All these races are kept 
in genetic continuity by interracial breeding 
and hybrid race types, so that the species, as 
a whole, has a single gene pool containing 
no portion completely isolated from any 
other. On the other hand, different cross- 
fertilizing species are genetically discontinu- 
ous from each other. Thus, the gene pool 
of one species is so isolated from the gene 
pools of all other species that none can lose 
its identity via crossbreeding, or hackcross- 
ing subsequent to crossbreeding. Moreover, 
the gene pools of different species are iso- 
lated from each other for genetic — not 
merely environmental — reasons. 

The formation of new species, speciation, 
has occurred frequently in past evolution; 
since evolution is continuing, new species 
are still being formed. The speciation mech- 
anism considered most common for cross- 
fertilizing individuals involves the production 
of two or more species from a single one. 
How can this come about? 

Hypothetically, one can start with a single 
panmictic, genetically-polymorphic species. 
Since environments vary we will assume that 
different populations occupy different por- 
tions of a territory and, although enough 
interpopulation breeding takes place to form 
one gene pool, most of the breeding is intra- 
population. If, in the course of time, two 
(or more) of these populations diverge ge- 
netically — each one uniquely adapted to its 
own territory — these populations become 
different races of the same species. The dif- 
ferences in the gene pools of these two races 
may increase more and more because of mu- 
tation, natural selection, and genetic drift. 
As this differentiation process continues, the 
genes which make each of the races adaptive 
in their own territories may, by their mani- 



fold phenotypic effects, make matings be- 
tween the two races still less likely to occur 
or may cause the hybrids of such matings 
to be less adaptive than the members of 
either parent race. Accordingly, partial re- 
productive isolation may be initially an acci- 
dental or an incidental byproduct of the 
adaptability of genotypes to a given en- 
vironment. The greater this effect, however, 
the greater we would expect the selective 
advantage to be of genes which increase the 
reproductive isolation between two diverging 
races further still. If races continued to di- 
verge genetically in this way, they would 
eventually form separate and different gene 
pools, and instead of being two races of the 
same species would become two different 
species. Note that speciation is an irrevers- 
ible process; once a gene pool has reached 
the species level, it can never lose its identity 
via cross breeding with another species. 

In this generalized account of how specia- 
tion usually occurs races have acted as in- 
cipient species. But remember that under 
other circumstances two races can also cross- 
breed to become a single race. For exam- 
ple, although several thousand years ago dif- 
ferent allopatric populations of human be- 
ings were definitely different races which 
might have formed different species had the 
same conditions of life continued, some of 
these races subsequently merged into one 
race because civilization and migration facili- 
tated crossbreeding. 

Gene exchange between races can be hin- 
dered in several ways. Those barriers lead- 
ing to complete reproductive isolation in- 
clude the following: 

1. Geographical. Water, ice, mountains, 
wind, earthquakes, and volcanic activity 
may separate races. 

2. Ecological. Changes in temperature, hu- 
midity, sunlight, food, predators, and 
parasites may alter or completely change 
a race's habitat. 



Races and the Origin of Species 



245 



3. Seasonal. Seasonal changes may cause 
different races to become fertile at differ- 
ent times even if their territories overlap, 
or if they are sympatric. 

4. Sexual or ethological. Intrarace mating, 
due to preference or domestication ef- 
fected by man. 

5. Morphological. Incompatibility of the 
sex organs between some races. 

6. Physiological. Failure of a race's sex 
cells to fertilize those of another, so that 
the hybrid zygote is formed infrequently, 
or not at all. 

7. Hybrid inviability. Even when formed, 
the development of hybrid zygotes may 
be so abnormal that it cannot be com- 
pleted. 

8. Hybrid sterility. A possibility even if 
hybrids complete development and are 
hardy. 

Although geographical, ecological, and sea- 
sonal differences do not automatically initi- 
ate genotypic differences, they furnish the 
environmental variations which select from 
the available genotypes those which are 
adaptive; that is, those with the greatest re- 
productive potential under the given condi- 
tions. Of course, mutation must provide 
the raw materials for natural selection; since 
no single genotype is equally well adapted 
to all conditions, different races come to 
contain different genotypes. The remaining 
barriers listed can complete reproductive 
isolation. 

The many genes by which two incipient 
species differ may produce seasonal, sexual, 
morphological, and physiological barriers. 
Hybrid inviability may result from develop- 
mental disharmony caused by the presence 
of two genetically different genomes in each 
cell. Although hybrid sterility can be caused 
by such genetic action, it also results when 
two races become quite different with respect 
to gene arrangement — because of structural 



changes within and between chromosomes — 
so that during meiosis, synapsis between the 
two different genomes in the hybrid is irreg- 
ular. Improper pairing causes abnormal 
segregation, which results in aneuploid meio- 
tic products. Recall that aneuploidy in pol- 
len is lethal, and that aneuploid gametes in 
animals usually result in dominant lethality 
of the zygotes they form. Consequently, 
reproductive isolation can be based upon 
either genetic activity or chromosomal be- 
havior, or both. 

It seems reasonable that the more morpho- 
logically divergent two forms are, the more 
likely it is that they will differ physiologically 
and that these differences will have orig- 
inated in very different and isolated gene 
pools. Simply by comparing horse and 
mouse morphologies, one certainly expects 
them to be different species; thus the occur- 
rence of morphological differences is some- 
times a good index of a species difference. 
However, when the groups being compared 
are closely related in descent, one finds that 
morphology is not well correlated with re- 
productive isolation. For example, Euro- 
pean cattle and the Tibetan yak are quite 
different in appearance and usually are 
placed in different genera, but these two 
species can be crossed. Moreover, in Tibet, 
many cattle have yak-like traits, so that 
widely different phenotypes do not neces- 
sarily result in complete reproductive isola- 
tion between closely related species. On the 
other hand, consider D. persimilis and D. 
pseudoobscura. These two species — for- 
merly considered races of the same species 
— are so similar morphologically that they 
can be differentiated by their genitalia only 
if very careful measurements are made. 
Nevertheless, these two species have com- 
pletely isolated gene pools in nature, even 
where their territories overlap. Such mor- 
phologically similar species are called sibling 
species. They originated from different 



246 



CHAPTER 18 



i aces of a single species. Sibling species are 
found in mosquitoes and Other insects as 
well as in Drosophila; they are also found 
in plants — among the tarweeds of the aster 

family and in the blue wild rye. 

The study of D. pseudoobscura and D. 

persimilis illustrates two other principles re- 
lating to species formation. First, any par- 
ticular reproductive barrier usually has a 
multigenic and or a multichromosomal basis; 
second, any two species are separated not 
by one but by a number of reproductive bar- 
riers. Although each of the barriers involved 
is incomplete, together they result in com- 
plete reproductive isolation — there being no 
stream of genes between the two gene pools 
in nature. The known differences between 
these two particular sibling species include: 

1 . Pseudoobscura lives in drier and warm- 
er habitats than persimilis 

2. Females accept the mating advances 
of males of their own species more 
often than they do male advance of 
the other 

3. Pseudoobscura usually mates in the 
evening, persimilis in the morning 

4. Interspecific hybrids are relatively in- 
viable and when viable, they are mostly 
sterile. 

The nature and origin of the reproductive 
isolation mechanisms involved in forming 
new species from races shows that valid 
species originate not by a single or simple 
mutation, but as the result of many different, 
independently occurring genetic changes. 
Moreover, as already noted, speciation is 
accomplished not merely by an accumulation 
of mutants which distinguish races, but also 
by those which contribute to reproductive 
isolation. Usually populations are physically 
separated while reproductive barriers are 
being built up; otherwise, hybridization 
would break down these barriers. Experi- 
mental evidence also supports our expecta- 



tion that natural selection acts to further the 
accumulation of the genetic factors promot- 
ing reproductive isolation between races. 

The preceding discussion illustrates how 
one species can give rise to two or more 
species via races which serve as incipient 
species.- It was stated earlier that a species 
has an isolated gene pool, that is, a gene 
pool closed to individuals of some other al- 
ternative condition (species). A species is 
expected to undergo numerous changes in 
its gene pool during the course of many 
generations. At the end of this time, is it 
the same or a new species? Here is an 
example of one type of species formation 
which would not be recognized by the cri- 
terion above because the alternative state 
would no longer exist. Suppose some mem- 
bers of the original population had been 
(miraculously) preserved, then we might 
find that they were reproductively isolated 
from the members of the new population. 
In such an event we could admit the forma- 
tion of a new species whose origin is de- 
pendent upon the "extinction" of the parent 
species. This type of speciation will be- 
come a valid subject of study once man 
learns how to preserve sample genotypes 
indefinitely. 

One species can give rise to another via 
allopolyploidy — an increase in the number 
of genomes present in a normally cross- 
fertilizing species. Mechanisms for the pro- 
duction of autopolyploid cells, tissues, and 
organisms have already been described on 
pp. 151-153. In the genus Chrysanthemum, 
species occur with 2n chromosome numbers 
of 18, 36, 54, 72, and 90. Thus, it appears 
that nine is the basic n number. In the 
genus Solanum (the nightshades, including 
the potato ) the basic n number seems to be 
twelve, since species of this genus are known 



'-' See Th. Dobzhansky, L. Ehrman, O. Pavlovsky, 
and B. Spassky (1964). 



Races and the Origin of Species 



247 



having 24, 36, 48, 60, 72, 96, 108, and 144 
chromosomes. These two examples sug- 
gest that autopolyploidy has played a role 
in the speciation of these two genera. Auto- 
polyploidy, however, is not considered an 
important mechanism of speciation in forms 
reproducing primarily by sexual means, 
since autopolyploids having more than 2n 
chromosomes form multivalents at meiosis 
and, therefore, numerous aneuploid gametes. 
Autopolyploids can succeed, though, if they 
are propagated asexually, by budding or 
grafting, as in the case of the triploid apples 
— Gravenstein and Baldwin. Triploid tulips 
are also propagated asexually. 

Speciation Involving Two or More Species 

Many new cross-fertilizing species originate 
not only from a single species or its races, 
but — in relatively recent times — from hy- 
bridization between two or more different 
species, that is, via interspecific hybridiza- 
tion. Although interspecific hybrids pose no 
threat to the isolation of the gene pools of 
their parental species, they may form a suc- 
cessful, sexually-reproducing population that 
has its own closed gene pool. Interspecific 
hybrids, particularly of plants, can be con- 
verted into stable, intermediate types iso- 
lated from their parental species by three 
methods. 

The first method involves amphiploidy 
(allopolyploidy, see p. 155). If one species 
has 2n = 4 and another has 2n = 6, the Fi 
hybrid between them will have five chromo- 
somes (Figure 18-2). If the hybrid sur- 
vives, it may be sterile because each chro- 
mosome has no homolog and, therefore, no 
partner at meiosis. As a result, meiosis pro- 
ceeds as if the organism were a haploid and 
produces mostly aneuploid gametes. If, 
however, the chromosome number of the 
Fi hybrid is doubled — either artificially (via 
colchicine) or spontaneously — the individual 
or sector will be 2n = 10; each chromosome 



2n 4 



2n 6 






Aneuploid 
-*■ Meiotic 
Products 



AMPHIPLOID 




Euploid(n) 
"*" Meiotic 
Products 



figure 18-2. Interspecific hybridization lead- 
ing to new species formation via amphiploidy 
(allopolyploidy). 



will have a meiotic partner; and euploid 
gametes of n = 5 will be formed. Upon 
uniting, such gametes produce 2n = 10 
progeny, which are fertile and more-or-less 
phenotypically intermediate to and isolated 
from both parental species. 

It has been estimated that twenty to 
twenty-five per cent of the present flowering 
plant species originated as interspecific hy- 
brids whose chromosomes doubled in num- 
ber (therefore being "doubled hybrids" or 
amphiploids). Moreover, in the past many 
(or more) species originated in this way, 
then diverged to form different genera. Nat- 
urally-occurring amphiploidy was involved 
in the origin of cotton in the New World 
and in the appearance of new species of 
goatsbeard during the present century. 

In the early 1800's, the American marsh 
grass, S parti na alterni flora (2n = 70), was 



2»s 



CHAPTER 18 




DIPLOID 
HYBRID 



AMPHIPLOID 



figure 18-3 {above). Seed pods of cabbage 
and radish, of their hybrid and amphiploid. 
{After G. D. Karpechenko.) 



figure 18-4 (below). Distribution of Del- 
phinium species in California. Each species 
has a unique habitat. 



D. GYPSOPHILUM 




D. RECURVATUM 
D. HESPERIUM 



accidentally transported by ship to France 
and England and became established along- 
side of the European marsh grass, 5". Stricta 
(2n 56). By the early 1900's a new 
marsh grass, S. townsendii (2n = 126), ap- 
peared and largely erowded out the two 
older species. Since S. townsendii has a 
chromosome number equal to the sum of 
the diploid numbers of the older species, is 
fertile, breeds true, and has an appearance 
intermediate between the two older forms, 
this species is undoubtedly an amphiploid 
of S. alterniflora and S. stricta. S. town- 
sendii is so hardy that it has been purposely 
introduced into Holland (to support the 
dikes) and to other localities. 

Amphiploidy also can be produced arti- 
ficially. For example, in the greenhouse 
it is possible to cross radish (2n = 18) with 
cabbage (2n = 18) (Figure 18-3), thus 
producing an F] hybrid with 18 unpaired 
chromosomes at meiosis. If, however, the 
chromosome number of the hybrid doubles 
early enough in development, it can produce 
amphiploid progeny with 2n = 36 chromo- 
somes (containing nine pairs each from 
radish and cabbage). Since the amphiploid 
is fertile and genetically isolated from both 
radish and cabbage, it constitutes a new 
species. 

If each chromosome contributed to an in- 
terspecific hybrid is different and the hy- 
brid's chromosome number doubles, then 
each chromosome would have just one part- 
ner at meiosis, and segregation would be 
normal. Therefore, the breeding success of 
the amphiploid is enhanced by greater dif- 
ferences between the chromosomes of the 
two species that contributed haploid ge- 
nomes to the interspecific hybrid. It is not 
surprising, then, in hybridizing two chromo- 
somally similar species, that at meiosis their 
amphiploid produces trivalents and quad- 
rivalents leading to abnormal segregation 
and sterility. 

Although amphiploidy is not successful 



Races and the Origin of Species 



249 



for hybrids between similar species, there is 
a second way interspecific hybrids can be- 
come stabilized as a new species, provided 
that the two hybridizing species are very 
similar chromosomally. If the two species 
have the same haploid number, their Fi 
hybrid may have all chromosomes synapsed 
in pairs at meiosis. Segregation, independ- 
ent segregation, and crossing over may yield 
progeny of the hybrid whose recombinations 
can become stabilized in nature yet are iso- 
lated from either parental species. Con- 
sider certain species in the larkspur genus, 
Delphinium: D. gypsophilum is morpholog- 
ically intermediate between D. recurvatum 
and D. hesperium; all three species have 
2n = 16; and the "parent" species, recurva- 
tum and hesperium, can be crossed to pro- 
duce an ¥\ hybrid. When the F! hybrid is 
crossed to gypsophilum, the offspring are 
more regular and more fertile than those 



produced by backcrossing the F 1 hybrid with 
either parent species. Similarly, the progeny 
from crosses between gypsophilum and either 
of its parent species are not as regular or as 
fertile as are those from the cross between 
gypsophilum and the hybrid of the parent 
species. These results provide good evi- 
dence that gypsophilum arose as the hybrid 
between recurvatum and hesperium. Fig- 
ure 1 8-4 shows the distribution of these 
species in California. 

The third way that interspecific hybrids 
can become stabilized as new species is by 
introgression. In this process a new type 
arises after the interspecific hybrid back- 
crosses with one of the parental types. The 
backcross recombinant types favored by nat- 
ural selection may contain some genetic com- 
ponents from both species, may be true- 
breeding and, eventually, may become a new 
species. 



SUMMARY AND CONCLUSIONS 

A race of a cross-fertilizing species is characterized by the content of its gene pool. 
Each race is adapted to the territory in which it lives. Different races can be sympatric 
or allopatric. Races can become species by accumulating genetic differences whose 
end effect is genetic discontinuity — that is, the formation of isolated gene pools. Sep- 
aration of two gene pools is usually accomplished by a combination of different repro- 
ductive barriers each of which is incomplete by itself and has a polygenic and/or a 
polychromosomal basis not necessarily correlated with morphological differences. 

It is generally recognized that most cross-fertilizing species arose from the further 
differentiation of races. Occasionally a new species can arise via autopolyploidy, and 
it is possible that a new species can also arise by the gradual change of one species as 
a whole into another species. 

Two (or more) species can give rise to a new one after interspecific hybridization. 
An interspecific hybrid can form a new species via amphiploidy, by selection of re- 
combinants among its progeny, or by selection of individuals produced after intro- 
gression. 



2"0 CHAPTER 18 

REFERENCES 

Dobzhansky, I h.. Genetics and the Origin of Species, 3rd Ed.. New York: Columbia 
University Press, 1951. 

Dobzhansky, Th., Evolution, Genetics, and Man, New York: John Wiley & Sons, 1955. 

Dobzhansky, In.. Ehrman, 1... Pavlovsky, O., and Spassky, B., "The Supcrspecics Dro- 
sophila paulistorum," Proc. Nat. Acad. Sci., U.S., 51:3-9, 1964. 

Dodson. E. O.. Evolution: Process and Product (Rev. Ed.), New York: Rinehart, 1960. 

Dunn. L. C. and Dobzhansky, Th., Heredity, Race, and Society, 3rd Ed., New York: 
New Amer. Libr. of World Lit.. 1957. 

Ehrlich. P. R.. and Holm. R. W.. The Process of Evolution, New York: McGraw-Hill 
Book Co., Inc.. 1963. 

Mayr, E., Animal Species and Evolution, Cambridge: Harvard University Press, 1963. 

Merrill. D. J.. Evolution and Genetics, New York: Holt, Rinehart and Winston, 1962. 

Stebbins, G. L., Variation and Evolution in Plants, New York: Columbia University 
Press, 1950. 



QUESTIONS FOR DISCUSSION 

18.1. Discuss the validity of the concept of a pure race. 

18.2. To use the frequencies of ABO blood types in tracing the course of past migra- 
tion, what assumptions must you make? 

18.3. Under what future circumstances would you expect the number of races of 
human beings to decrease? To increase? 

18.4. Can the definition we have used for a species be applied to forms that reproduce 
only asexually? Why? 

18.5. Differentiate between genetic sterility and chromosomal sterility. Invent an 
example of each type. 

18.6. Discuss the hypothesis that a new species can result from the occurrence of a 
single mutational event. 

18.7. Is geographical isolation a prerequisite for the formation of a new species? 
Explain. 

18.8. What is the relative importance of mutation and genetic recombination in species 
formation? 

18.9. Is a species a natural biological entity, or is it — like a race — defined to suit 
man's convenience? 

18.10. Does the statement, "We are all members of the human race," make biological 
sense? Why? 

18.11. Suppose intelligent beings, phenotypically indistinguishable from man, arrived 
on Earth from another planet. Would intermarriage with Earth people be likely 
to produce fertile offspring? Why? 

18.12. Invent circumstances under which the present single species of man could evolve 
into two or more species. 

18.13. The cells of triploid and tetraploid autopolyploids are usually larger than those 
of the diploid. What importance has this fact for fruit growers? 



Races and the Origin of Species 251 

18.14. H. Kihara and co-workers have produced triploid (33 chromosomes) water- 
melons with no seeds, and tetraploid (44 chromosomes) watermelons with seeds 
but larger than the diploid. How do you suppose this was accomplished? How 
do you suppose these types are maintained? 

18.15. In each of the following cases, an interspecific hybrid can be formed experi- 
mentally. State whether or not you would expect each of the hybrids produced 
from the parents described below to become established in nature. 

(a) In California, the Monterey cypress grows along the coast on the rocks. 
whereas the allopatric Gowen cypress grows two miles inland in the sand 
barrens. 

(b) Two sympatric species of pine occur in California. One of these, the 
Monterey pine, sheds its pollen before March; the other, the bishop pine, 
some time later. 

(c) The hybrid between Crepis neglecta (n = 4) and C. fuliginosa (n = 3) 
shows unpaired and paired chromosomes as well as multivalents during 
meiosis. 

18.16. Drosophila pseudoobscura crossed with D. persimilis produces sterile males, but 
partially fertile females. Using marked chromosomes, the interspecific hybrid 
female can be backcrossed to pseudoobscura, and their progeny can have various 
combinations of the chromosomes of the two species. When male progeny of 
the backcross are examined for the length of the testis, it is found that the testis 
is essentially normal when the X chromosome is from pseudoobscura; when the 
X is from persimilis, the testis is shorter — increasing in abnormality as the num- 
ber of autosomes coming from pseudoobscura increases. What can you con- 
clude about reproductive barriers from these results? 

18.17. The cotton species Gossypium hirsutum and G. barbadense are tetraploids 
(2n = 52) and are phenotypically intermediate between the diploid species G. 
herbaceum and G. raimondii. (Each is 2n = 26.) If cytological examination 
is made of meiosis in various hybrids, what would the following results reveal 
about the origin of these species? 

(a) barbadense X raimondii shows 13 pairs and 13 singles. 

(b) barbadense X herbaceum shows 13 pairs and 13 singles. 

(c) raimondii X herbaceum shows 26 singles. 

18.18. What do the following two cases have in common? 

(a) Through artificial selection the evolution of corn, Zea mays, was aided by 
genes incorporated from teosinte, Zea mexicana. 

(b) Commercial wheat contains genes for rust resistance obtained from goat 
grass chromosomes. 



Chapter 19 

CHEMICAL NATURE OF GENES 



I 



n the preceding chapters the pri- 
mary concern was with the defi- 
nition of the genetic material on 
the basis of its capacity to recombine and 
mutate; our concern here will be with the 
chemical nature of the genetic material as 
revealed through chemical analyses. Let us 
try to determine which of the cell's chemical 
components are and which are not suitable 
to serve as genetic material. Since the nu- 
cleus contains genetic material in its chro- 
mosomes, any chemical substances located 
exclusively in the cytoplasm can, of course, 
be eliminated from consideration as the basis 
for nuclear genetic material. Because the 
genetic material seems to possess complex 
properties, one would expect that its chemical 
properties were also complex. On this basis, 
we can eliminate from consideration all in- 
organic compounds (compounds not con- 
taining carbon), since no class of inorganic 
compound enters into a sufficient variety of 
chemical reactions. 

One unique feature of protoplasm is the 
speed and orderliness of its chemical activi- 
ties. These two characteristics are due to 
the presence of proteins in the form of en- 
zymes and cellular structures. Different 
kinds of proteins contain different numbers 
of amino acids. Since twenty or so different 
kinds of amino acids are found in the pro- 
tein of organisms, the total number of dif- 
ferent combinations is, for all practical pur- 
poses, infinite. Protein clearly possesses 
adequate complexity, so it is not unreason- 
252 



able to hypothesize that the genetic material 
is composed of protein. 

It' the gene were protein in nature, one 
would expect to find protein in the chromo- 
somes but not, perhaps, of a type usually 
found in the cytoplasm. Chemical analyses 
of nuclei and chromosomes confirm both ex- 
pectations by revealing the existence of 
histories — complex proteins which act as 
bases and are found primarily in chromo- 
somes. Although the chromosomes of many 
cells contain histones, they are not found 
in the chromosomes of all cells. For ex- 
ample, histones are usually present in the 
somatic nuclei of fish; however, the sperm 
of trout, salmon, sturgeon, and herring in- 
stead contains protamine, a basic protein of 
simpler composition. The protamine in fish 
sperm is in turn replaced by histone in the 
somatic cells produced mitotically after fer- 
tilization. If genetic material is protein, the 
genetic specifications or information must be 
transferred from protamine to histone to 
protamine. At least in some organisms, 
then, the same genetic specifications would 
have to be carried in two chemical forms, 
protamine and histone. 

Present knowledge does not prevent us 
from entertaining the view that alternative 
chemical compositions are possible for the 
genetic material, but any alternative must 
be capable of performing a number of activi- 
ties in accordance with the principles already 
established. Nevertheless, the hypothesis 
that protamine and histone are both genetic 
material in the same organism is rather com- 
plicated, at least when one considers that 
there would be two chemical formulae for 
a single genotype. For the sake of sim- 
plicity, it would be more satisfactory if a 
single nuclear chemical substance were the 
genetic material. 

Other proteins are found in chromosomes. 
Their quantity changes, however, according 
to the type and rate of metabolic activity 



Chemical Nature of Genes 



253 



performed by the cell. There is, therefore, 
no simple one-to-one relationship between 
their quantity and gene quantity. Conse- 
quently, additional hypotheses are required 
to explain genetic behavior. Despite the 
initial attractiveness of the hypothesis that 
the genetic material is proteinaceous, one 
can conclude that the types and amounts of 
nuclear protein actually found do not ade- 
quately support this view. 

Chromosomes contain another chemical 
substance which seems to be absent in the 
cytoplasm (Figure 19-1). This chemical 
is a type of nucleic acid 1 called deoxyribo- 
nucleic acid, or DNA, a substance usually 
found combined with basic proteins like 
protamine and histone (by means of a chem- 
ical linkage not completely understood) to 
form deoxyribonucleoproteins. Before in- 
vestigating the possibility that chromosomal 
DNA is genetic material, let us consider first 
the chemical composition of its molecule. 

Chemical Composition of DNA 

Organic bases. Chromosomal DNA con- 
tains organic ring compounds of which nitro- 
gen is an integral part. The fundamental 
N-containing structure is a six-membered 
ring, as found in benzene, C ( ;H (i . Figure 
19-2a shows the complete structural ar- 
rangement of benzene; Figure 19-2a' is an 
abbreviated version with the carbon atoms 
of the ring omitted; Figure 19-2a" is the 
same model condensed further by eliminat- 
ing the hydrogen atoms attached to ring 
carbon atoms. The basic N-containing ring 
in DNA is a pyrimidine . This molecule 
(Figure 19-2b) has N substituted for the 
CH group at position 1 as well as at position 
3 in the benzene ring. Figure 19-2b' and 
Figure 19-2b" show successive abbrevia- 
tions of this formula corresponding to those 
used for benzene. 

1 Discovered by F. Miescher (1869). 



The nitrogen found in DNA is also found 
in a derivative of the basic pyrimidine ring, 
called a purine. This molecule consists of 
a pyrimidine ring — minus the H atoms at 
positions 4 and 5 — to which an imidazole 
ring (5-membered) is joined, so that the 
carbons at these positions are shared by both 
rings (Figures 19-2c, c', and c"). Hence- 
forth, the most abbreviated structural repre- 
sentation will be used for pyrimidines and 
purines. Since all pyrimidines and purines 
act chemically as bases, they are termed 
organic bases. 



*r 




••'• 




• *«• 




x»\ 


9 






IXx 









"•*y.i 




u * 



figure 19-1. Whole mount of a larval salivary 
"land of Drosophila. DNA stain is restricted 
to the nuclei. {Courtesy of J. Schultz.) 



2.")4 



CHAPTER 19 



Figure 19—3 shows the structural formulae 
for various types of pyrimidine. The names 
underlined are found in DNA. All the dc- 
rivatives shown o\' the basic pyrimidine ring 
have an oxygen at position 2 replacing the 
H which is now at position 3. This oxygen 

R 
is shown in the keto form (O = C<^ , with 

R 

R representing an atom or group other than 



H). Two pyrimidines are commonly found 
in DNA: cytosine and thymine. Cytosine 
differs from the basic pyrimidine ring by 
having an amino (NH L .) group attached to 
the C at position 6 instead of the H. Conse- 
quently, cytosine can also be called 6-amino- 
2-oxypyrimidine. Replacement of the H at 
position 5 in cytosine by a methyl (CH. { ) 
group produces 5-methyl cytosine; this DNA 
pyrimidine is found in appreciable amounts 



H 

•is 

H— Ci sC— H 

I II 

H— Q , 4C— H 



H 

a 




a' 
BENZENE 




H 

Ni sC— H 

I II 

H— Q , 4C— H 




b' 
PYRIMIDINE 




b" 



H 

Ni C 
H— O C 

H 



*C— H 





H 



PURINE 

[KiLRi 19-2. Relationship between certain ring compounds. 



Chemical Nature of Genes 



255 



Pyrimidine I 5 

N 




NH, 




(6-amino-2-oxy pyrimidine) 



NH, 




NH. 



CH, 




CH 2 OH 



5-METHYL CYTOSINE 
(6-amino-2-oxy-5- 
methylpyrimidine) 



5-HYDROXYMETHYL CYTOSINE 

(6-amino-2-oxy-5-hydroxy- 

methyl pyrimidine) 








O 


H— Nf^ji 




H-Nf^jpCH 3 


1 




1 


H 




1 
H 


URACIL 




THYMINE 


(2,6-oxypyrimidine) 


(2,6- 


-oxy-5-methyl pyrimidine 
(5-methyl uracil) 



figure 19-3. Pyrimi- 

dines. Names of pyrimi- 

dines occurring in DNA 

are underlined. 



in wheat germ and in trace amounts in mam- 
mals, fish, and insects. Another pyrimidine, 
found only in the DNA of certain viruses 
attacking bacteria, has a hydroxymethyl 
(CH.OH) group replacing the H at position 
5 of cytosine. and is therefore called 5 -hy- 
droxy methyl cytosine. 

The other pyrimidine commonly found in 
DNA is thymine. Thymine is unique in 
having a keto group replace the H attached 
to the C at position 6; in addition a methyl 
group replaces the H at position 5. So 
thymine can also be called 2,6-oxy-5-methyl- 
pyrimidine. Note that all the pyrimidines 
shown differ primarily in the groups present 
at the 5 and 6 positions in the ring. 



Figure 19-4 shows the structural formulae 
for various purines; those found in DNA are 
underlined. Two purines are commonly 
found in DNA: adenine and guanine. Ade- 
nine differs from the basic formula of purine 
by having an NH. group in place of H at 
position 6; therefore this compound can also 
be identified as 6-amino-purine. A deriva- 
tive of adenine has a CH ;! substitute for an 
H in the NH 2 group at position 6 with 6- 
methylaminopurine resulting; limited amounts 
of this purine have been found in DNA. 

The other purine most frequent in DNA 
is guanine (Figure 19-4). Since guanine 
has an NH 2 group at position 2 and an O 
in keto form at position 6, it can also be 



256 



CHAPTER 19 



figure 19-4. Purines. Names oj 
purines occurring in DNA arc un- 
derlined. 



Purine 




NH 




CH 


3 


N- 


H 


N^^ 




:> 

N 




H 


6-METHYLAMINOPURINE 



NH 




(6-aminopurine) 



Cm NT N 



2-METHYL ADENINE 
(2-methyl-6-aminopurine) 



CH, 
N— CH 3 

^N" "N 
I 
H 

6-DIMETHYLAMINOPURINE 





GUANINE 
[2-amino-6-oxy purine) 




N 
I 
H H 

2-METHYLAMINO GUANINE 




1 -METHYL GUANINE 



Chemical Nature of Genes 



257 



OH 




OH 



OH 




or 




OH 



D-RIBOSE 



or 




OH 



2-DEOXY-D-RIBOSE 



figure 19-5. Pentoses found in nucleic acids. 



called 2-amino-6-oxypurine. The purines 
differ largely in the groups attached at the 
2 and 6 positions of the double ring. 

Pentoses. D-ribose is a sugar (Figure 
19-5 a) containing five carbons (being, 
therefore, a pentose), four of which are 
joined with an O to form a five-membered 
ring. Figure 19-5 a' employs the conven- 
tion, used hereafter, of not showing the car- 
bons of the ring. The carbons in pentose 
are given primed numbers to indicate their 
positions. DNA contains a pentose modified 
from the D-ribose structure by the absence 
of an oxygen at position 2', so that this sugar 
is named 2' -deoxy -D-ribose and is often 
called 2-deoxyribose or simply, deoxyribose 
(Figures 19-5b and b')- 

Deoxyribosides. Each purine or pyrimi- 



dine base in DNA is joined to a deoxyribose 
sugar to form the combination called a de- 
oxyribonucleoside or deoxy riboside. The 
four main deoxyribosides in DNA are: de- 
oxy cytidine (for cytosine); (deoxy) thy- 
midine (for thymine); deoxy adenosine (for 
adenine); and deoxyguanosine (for gua- 
nine). The structure for these is shown in 
Figure 19-6. Note that the deoxyribose 
always joins to the organic base at its Y 
position. The linkage involved occurs at 
position 3 in pyrimidines and at position 9 
in purines. 

Deoxyribotides. In DNA a phosphate 
(PO^ group is always joined to a deoxy- 
riboside, forming a deoxyribonucleotide or 
deoxyribotide. The phosphate is attached 
either at the 3' or 5' position of the sugar 



:.-»s 



CHAPTER 19 



I li.IRE 19-6. 

Common deoxy 'ribosides. 



OH 



NH 



NP 5il 



.A 



O N 





OH 



CH 



OH H 

Deoxycytidine 



OH H 

Thymidine 



PYRIMIDINE DEOXYRIBOSIDES 



NH, 




OH H 

Deoxyadenosine 




OH 



CH 



OH H 

Deoxyguanosine 



PURINE DEOXYRIBOSIDES 



figure 19-7. Deoxyrihotides. 



OH 



5-CH 




Purine 

or 

Pyrimidine 




Deoxyriboside 3 -monophosphate 



OH 



Deoxyriboside 5 -monophosphate 



Chemical Nature of Genes 



259 



(a generalized form appears in Figure 19-7 ). 
This combination is shown specifically for 
the deoxyribotides containing the pyrimidine 
cytosine and the purine adenine in Figure 
19-8. The deoxyriboside 5'-monophos- 
phates of cytosine. thymine, adenine, and 
guanine are called, respectively, deoxycyti- 
dylic acid, thymidylic acid, deoxyadenylic 
acid, and deoxyguanylic acid. In summary, 
then, the basic unit of DNA is the deoxy- 



ribotide which is composed of a phosphate 
joined to a deoxyriboside; this, in turn, is 
composed of a deoxyribose joined to an or- 
ganic base. These bases are either pyrimi- 
dines (most commonly cytosine and thy- 
mine) or purines (most commonly adenine 
and guanine). 

Poly deoxyribotides. Chromosomal DNA 
occurs not as single deoxyribotides but as 
polydeoxyribonucleotides or polydeoxyribo- 




NHo 




Deoxycytidine 3 -monophosphate 



OH H 



Deoxycytidine 5'-monophosphate 

or 

Deoxycytidylic acid 



NH. 



NH 





Deoxyadenosine 3 -monophosphate 

figure 19-8. Specific deoxyribotides 



OH 



Deoxyadenosine 5 -monophosphate 

or 

Deoxyadenylic acid 



260 



CHAPTER 19 



tides. These molecules are actually chains 
in which the individual deoxyribotides com- 
prise the links. The way these links are 
joined can be understood by examining the 
two separate deoxyriboside 5'-monophos- 
phates shown at the right of Figure 19-8. 
I hese two compounds can become linked 
if the topmost () of the lower compound 
replaces the OH at position 3' of the sugar 




* Pyrimidine or purine base of 
appropriate type (usually cyto- 
sine. thymine, adenine or gua- 
nine). 

figure 19-9. Polydeoxyribotide. 



in the upper compound. (The same reac- 
tion occurs when a phosphate is added to 
position 3' of a deoxyriboside to produce a 
deoxyriboside 3'-monophosphate as illus- 
trated in the two molecules shown at the 
left of Figure 19—8.) Since deoxyriboside 
^'-monophosphates are capable of joining 
together by means of phosphate linkage at 
3', polydeoxyribotide chains of great length 
are produced. Figure 19-9 shows a portion 
of such a chain. Note that the polydeoxy- 
ribotide is a linear — that is, unbranched — 
molecule, whose backbone consists of sugar- 
phosphate linkages and whose linearity is in- 
dependent of the particular bases present at 
any point. This independence means that 
the structure of the chain is uninfluenced by 
the sequences of bases which can be in any 
array. Notice, moreover, that this polymer 
(a molecule composed of a number of iden- 
tical units) of deoxyribotides does not read 
the same in both directions. As indicated 
by the arrows, the sugar linkages to phos- 
phate read 3'5', 3'5', and so on; whereas in 
the opposite direction they read 5'3', 5'3', 
et cetera. Because of this difference, the 
polymerized DNA molecule is said to be 
polarized. 

Measuring DNA Quantity 

Two main methods are commonly used in 
determining the amount of DNA present in 
the nucleus: the histochemical and the cyto- 
chemical. The histochemical method em- 
ploys whole tissues for the chemical extrac- 
tion and measurement of DNA. Sometimes 
chemical analysis is made of masses of 
nuclei, from which most of the adhering 
cytoplasm has been removed by special treat- 
ment, to determine the average amount of 
DNA per nucleus. In the second, cytochem- 
ical, approach the DNA content of single 
nuclei, chromosomes, or chromosomal parts 
is determined. This method is based upon 
the finding that DNA is the only substance 



Chemical Nature of Genes 



261 



in the cell which stains under certain condi- 
tions. The Feulgen technique stains DNA 
purple (see p. 8), whereas the methyl green 
method stains DNA green. When properly 
applied, not only are these stains specific 
for DNA, but the amount of stain retained 
is directly proportional to the amount 
of DNA present. A given amount of dye 
bound by DNA will make a quantitative 
change in the amount of light of differ- 
ent wavelengths it transmits. These meas- 
urements can then be used to calculate the 
amount of DNA present. For example: a 
stained nucleus is placed under the micro- 
scope; different appropriate wavelengths in 
the visible spectrum are sent through the 
nucleus, and a series of photographs is taken; 
its DNA content is measured by density 
changes of the nucleus. From the italicized 
portions of words comes the name of this 
procedure, microspectrophotometry . 

A different application of microspectro- 
photometry utilizes another property of the 
purines and pyrimidines in DNA. These 
bases absorb ultraviolet light of wavelengths 
near 2600 A {Angstrom units). When 
other substances absorbing ultraviolet of 
these wavelengths are removed by enzymatic 
or other treatments, the quantity of DNA 
can be measured by its absorbence. As one 
test of the validity of the absorbency, one 
can remove the DNA from the chromosome 
by the use of enzymes — deoxyribonucleases , 
DNA uses, DNAses, or DNases. These or- 
ganic catalysts break the long DNA chains 
into short pieces which then can be washed 
out of the chromosomes and the nucleus. 
Such treatment produces the expected loss 
of absorbency. 

DNA as Genetic Material 

Having described the chemical nature and 
quantitative measurement of chromosomal 
DNA, we are in a position to consider some 
results bearing upon the relationship be- 



tween chromosomal DNA and the genetic 
material of the nucleus: 

1 . The quantity of DNA increases during 
the metabolic stage until it is exactly 
double (within the limits of experimental 
error) the amount present at the begin- 
ning of this stage. Mitosis apparently 
distributes equal amounts of DNA to the 
two telophase nuclei. Therefore, when 
first formed, all the diploid nuclei of an 
individual have approximately the same 
DNA content. 

2. The amount of DNA in a haploid gamete 
is roughly half that found in a newly 
formed diploid metabolic nucleus of the 
same individual. Fertilization, which re- 
stores the diploid chromosome condition, 
also restores the DNA content character- 
istic of the diploid cell. 

3. Polyploid cells increase proportionally in 
DNA content. 

4. Different cells of a tissue such as the 
salivary gland of larval Drosophila show 
different amounts of polynemy in their 
chromosomes. Since the DNA content 
of these different nuclei is found to be 
proportional to their volume, it is as- 
sumed to be a direct reflection of the 
degree of polynemy. 

5. The capacity of different wavelengths of 
ultraviolet light to induce mutations in 
fungi, corn, Drosophila, and other organ- 
isms is paralleled by the capacity of DNA 
to absorb these wavelengths. 

6. By tagging or labeling atoms (those that 
are radioactive or have an abnormal 
weight), it is found that many cellular 
components are being replaced continu- 
ously during metabolism. Despite this 
"atomic turnover," however, the total 
amount of cellular material does not in- 
crease. DNA is unusual because it shows 
little, if any, turnover; in other words, 
DNA maintains its integrity at the molec- 
ular level. 



262 



CHAPTER 19 



NUCLEIC 


COMMON 


PENTOSE 


NUCLEOSIDE 


(MONO-) NUCLEOTIDE 


ACID 


PYRIMIDINE (PY) 






with PO.at 5' 
4 




or PURINE (PU) 








BASE 












2 -deoxy-D-ribose 


deoxyriboside 


deoxyribotide 




Cytosinc PY 




Deoxycytidine 


Deoxycytidylic acid 




Thymine PY 




Thymidine 


Thymidylic acid 


DNA 












Adenine PU 




Deoxyadenosine 


Deoxyadenylic acid 




Guanine PU 




Deoxyguanosine 


Deoxyguanylic acid 






D - ribose 


riboside 


ribotide 




Cytosine PY 




Cytidine 


5 Cytidylic acid 




Uracil PY 




Uridine 


5 Uridylic acid 


RNA 












Adenine PU 




Adenosine 


5 Adenylic acid 




Guanine PU 




Guanosine 


5 Guanylic acid 



figure 19-10. Terminology for nucleic acids and their components. 



7. DNA is a linear, unbranched, polymer — 
a reasonable finding if one expected DNA 
to represent a sequence of genes. Just 
as interstitial genes are bipolar (see p. 
1 89 ) , so are interstitial segments of DNA, 
since each deoxyriboside joins only to 
two other deoxyribosides via its 3' and 5' 
sugar linkages to phosphate. 

In its cellular location and in all of the re- 
spects mentioned above, the observations 
are consistent with the view that DNA either 
is, or is intimately associated with, the ge- 
netic material. 

Chemical Composition of RNA 

In addition to DNA another type of nucleic- 
acid is found in the chromosome. This is 
ribonucleic acid or RNA. Normally chro- 
mosomal RNA is found in combination with 
protein in the form of rihonucleoprotein. 



Because the RNA content of chromosomes 
varies within a cell and among diploid cells 
of the same organism according to meta- 
bolic activity, RNA is unlikely to be the 
chemical basis of genes in typical (DNA- 
containing) chromosomes. Nevertheless, 
let us discuss the chemical composition of 
RNA, noting in particular its resemblance 
to DNA. 

Chromosomal RNA, like DNA, is a long, 
unbranched polymer with the basic unit being 
a ribonucleotide or ribotide. The ribotide 
is like the deoxyribotide, for it, too, is a 
combination of organic base plus pentose 
plus phosphate; one way in which it differs 
is that the pentose is D-ribose (Figure 19-5) 
rather than 2'-deoxy-D-ribose. Another dif- 
ference is found in RNA's pyrimidines. The 
two pyrimidines commonly found in RNA 
are cytosine (also common in DNA) and 
uracil (2,6-oxypyrimidine — -not found in 



Chemical Nature of Genes 



263 



typical DNA). Uracil's structure is shown 
in Figure 19-3. The two purines commonly 
found in DNA, adenine and guanine, are 
also common in ribotides. In RNA the base 
plus sugar combination is called a ribonu- 
cleoside or riboside. Ribosides are joined 
together by phosphates joined both at the 
3' and 5' positions of the sugar just as in 
DNA; consequently Figure 19-9 can repre- 
sent a polyribotide if an O is added at each 
2' position (making each sugar D-ribose), 
and if among the bases usually present uracil 
is substituted for thymine. It should be 
noted that RNA also absorbs ultraviolet 
light of 2600 A but can be removed from 
the chromosome by treatment with ribo- 
nucleases or RNases. 

In summary, typical chromosomes con- 
tain two nucleic acids, DNA and RNA. 
These normally occur in combination with 
protein to form nucleoproteins (deoxyribo- 
nucleoprotein and ribonucleoprotein) in 
which DNA and RNA occur as polynucleo- 
tides (polydeoxyribotides and polyribotides). 
Each polynucleotide is built of (mono-) 
nucleotides (deoxy- and ribotides, respec- 



tively), which in turn are composed of phos- 
phates joined at 5' of nucleosides (deoxy- 
ribo- and ribosides ) . These nucleosides are 
made up of a pentose (2'-deoxy-D-ribose 
and D-ribose) joined to a pyrimidine (usu- 
ally cytosine or thymine and cytosine or 
uracil) or to a purine (usually adenine or 
guanine). A portion of this terminology is 
summarized in Figure 19-10. 

Although the RNA in chromosomes pos- 
sesses neither the proper quantitative varia- 
tion nor the constancy expected of ordinary 
chromosomal genes, it does have the same 
linear organization as DNA. Moreover, 
some viruses composed primarily of ribo- 
nucleoprotein (influenza, poliomyelitis, and 
other encephalitic viruses; plant-attacking 
viruses such as the tobacco mosaic virus; and 
certain bacteria-attacking viruses) possess 
genetic properties but do not contain DNA. 
Since DNA rather than protein is favored 
as being the genetic chemical under typical 
chromosomal conditions, it is reasonable to 
consider RNA rather than the protein to be 
the chemical basis of genetic specification in 
these particular viruses. 



SUMMARY AND CONCLUSIONS 

This chapter is an initial attempt to throw some light on the chemical nature of the 
genetic material. The search for chemical substances with properties of the genetic 
material has led to a consideration of the protein found in chromosomes, but the avail- 
able evidence does not actively support such a primary role for protein. 

It is hypothesized that DNA either is or, at least, is intimately associated with the 
genetic material in chromosomes in view of the following: the localization of DNA; 
its quantity and distribution in mitosis, meiosis, and fertilization; its quantity in poly- 
ploid and polynemic situations; the parallel between DNA absorption and the muta- 
genicity of ultraviolet light; the maintenance of molecular integrity; and its long, linear, 
unbranched arrangement. It is also hypothesized that RNA may assume the genetic 
role in certain DNA-free viruses. Some details of the chemical nature of DNA and 
RNA are presented. 

Subsequent chapters will aim to further test the hypothesis that DNA (and RNA 
in special cases) is typically either the genetic material or intimately associated with 
it. Our ultimate objective is to determine the chemical units of the genetic material — 
chemical units corresponding to the genetic units of replication, mutation, recombina- 
tion, and function. 



264 CHAPTER 19 

REFERENCES 

Chargaff, E., and Davidson, J. N. (Eds.), The Nucleic Acids, 2 Vols., New York: 
Academic Press, 1955; Vol. 3, New York: Academic Press, 1960. 

Davidson. J. N.. and C'ohn, W. E. (Eds.), Progress in Nucleic Acid Research, 2 Vols., 
New York: Academic Press. 1963. 

Miescher. F.. "On the Chemical Composition of Pus Cells," translated in Great Experi- 
ments in Biology, Gabriel, M. L., and S. Fogel (Eds.), Englewood Cliffs, N.J.: 
Prentice-Hall. 1955. pp. 233-239. 

Potter. V. R.. Nucleic Acid Outlines, Vol. I. Minneapolis: Burgess Publ. Co., 1960. 

Steiner, R. F.. and Beers, R. F., Jr., Polynucleotides. Natural and Synthetic Nucleic- 
Acids, Amsterdam: Elsevier Publ. Co., 1961. 



QUESTIONS FOR DISCUSSION 

19.1. Is it simpler to postulate that DNA rather than protein constitutes the genetic 
material? Why? 

19.2. What is the chemical distinction between: 

(a) a mononucleotide and a polynucleotide? 

(b) a nucleotide and a nucleoside? 

(c) a pyrimidine and a purine? 

(d) a ribose and a deoxyribose sugar? 

19.3. Draw the complete structural formula of a polyribotide having the base sequence 
adenine, uracil, guanine, cytosine. 

19.4. Express thymine as a derivative of uracil. What part of the term deoxythymidine 
is superfluous? Why? 

19.5. What evidence can you provide to support the view that viruses possess genetic 
properties? 

19.6. How would you proceed to measure the absorbency of ultraviolet light by chro- 
mosomal DNA? Chromosomal RNA? 

19.7. Do you believe that the evidence so far presented provides conclusive proof 
that DNA is genetic material in chromosomes? Why? 

19.8. What is your opinion of the hypothesis that DNA is the chemical basis for re- 
combination, but that protein is the chemical basis for gene function? 

19.9. Is DNA complex enough to serve as the chemical basis of gene action? Explain. 

19.10. Do you think the term chemon could be defined usefully? Justify your opinion. 

19.11. Discuss the similarities and the differences between DNA and RNA. 



Chapter 20 

ORGANIZATION AND REPLICATION 
OF DNA IN VIVO 



T: 



|hat DNA serves as the chem- 
ical basis of chromosomal ge- 
netic material was supported 
by the indirect evidence presented in the 
last chapter. The primary structure of DNA 
was described as a single, long, unbranched, 
polarized chain of nucleotides. If the DNA 
polymer were genetic material, one would 
expect it to be linearly differentiated so that 
successive portions could represent different 
genes. This differentiation cannot be based 
upon either the deoxyribose sugar or the 
phosphate, since one of each is present in 
every nucleotide. Therefore, all differences 
in genetic information along the length of 
the DNA strand would have to be due to 
the organic bases present. Since species 
differ genetically, one might expect them to 
differ in DNA quantity and/or base content. 
Figure 20-1 gives the per genome DNA 
content of various types of organisms. It is 
generally true that the higher an organism 
is on the evolutionary scale, the larger is its 
genomic DNA content. Perhaps it would 
be more meaningful to say that the DNA 
content per genome increases as the number 
of functions controlled by genes increases. 
Histochemical analyses reveal the organic 
base content in DNA extracted from various 
species. Considering the total amount of 
the bases in an extract as 100%, we see in 
Figure 20-2 the portions found as adenine 
(A), thymine (T), guanine (G), and cyto- 
sine (C). There is considerable variation 
in the relative frequency of bases, ranging 
265 



from organisms relatively rich in A and T 
and poor in C and G (sea urchin) to those 
in which A and T are much less abundant 
than C and G (tubercle bacillus). The 
DNA samples taken from radically different 
species contained relatively different amounts 
of the four bases. 

Do these data suggest that a shift in the 
sequence of bases can produce genetic dif- 
ferences? The assumption that different 
orders of the same bases might be involved 
in specifying different genetic units is con- 
sistent with the fact that the chicken, salmon, 
and locust — certainly all very different ge- 
netically — have very similar base ratios. An 
alternative explanation would be that these 
species are molecular polyploids which differ 
only in the multiples of a basic set of DNA 
molecules they contain. This possibility can 
be eliminated from serious consideration in 
light of our knowledge that chromosomal 
polyploidy has made a limited contribution 
to evolution, at least in the animal kingdom 
(Chapters 11, 18). 

As long as relatively crude histochemical 
analyses are made of the total amount of 
DNA in cells with a large DNA content, 
we should expect to find roughly the same 
base ratios among the different members of 
a single species. This expectation has 
proved true. Moreover, the same base ratios 
are found in different normal and neoplastic 



Man, Mouse, Maize 


5-7 x 10 9 


Drosophila 


8x 10 7 


Aspergillus 


4x 10 7 


Escherichia 


1 x 10 7 


Bacteriophage T4 


2 x 10 5 


Bacteriophage XI 74 


4.5 x 10 3 (Unpaired) 



figure 20-1. DNA nucleotide pairs per ge- 
nome in various organisms. 



266 



( II A I'll R 20 



ADENINE THYMINE GUANINE CYTOSINE 



Man (sperm! 


1 


31.0 


31.5 


19.1 


18.4 


Chicken 




28.8 


29.2 


20.5 


21.5 


Salmon 




29.7 


29.1 


20.8 


20.4 


Locust 




29.3 


29.3 


20.5 


20.7 


Sea urchin 




32.8 


32.1 


17.7 


17.7 


Yeast 




31.7 


32.6 


18.8 


17.4 


Tuberculosis 


bacillus 


15.1 


14.6 


34.9 


35.4 


Escherichia coli 


26.1 


23.9 


24.9 


25.1 


Vaccinia virus 


29.5 


29.9 


20.6 


20.3 



E. coli bacteriophage T 2 32.6 



32.6 



18.2 



16.6 



figure 20-2. Base composition of DNA from various organisms. (*5-/;v- 
droxymethyl cytosine. ) 



tissues in the same and among different 

human beings. Nevertheless, a genome is 

expected to contain many DNA molecules 

which differ in base content and sequence. 

A _l_ t 

The variation found in — — in differ- 

G + C 

ent species — the ratio is about 0.4 for the 

tubercle bacillus and about 1 .8 for the sea 

urchin — is consistent with our chemical 

knowledge, since the DNA strand imposes 

no limitation upon either the types or the 

frequencies of the bases present along the 

length of the fiber. However, the amount 

of A and the amount of T in the DNA of 

a given species are remarkably equal as are 

the amounts of G and C (Figure 20-2). 

Since in each species A = T and G = C, 

it is also true that A + G = T + C;in other 

words, the total number of DNA purines 



always equals the total number of DNA 
pyrimidines. Although this regularity is 
common to all the chromosomal DNA's 
listed, there is nothing in the primary struc- 
ture of DNA which helps to explain this 
significant fact. That the primary structure 
of DNA is the same in all these organisms 
suggests, however, that these regularities 
may be connected with some additional, 
general characteristic of chromosomal DNA 
structure. 

An understanding of the basis for the 
A = T and G = C relationships may come 
from studies of an entirely different kind. 
It has been known for some time that a 
beam of X rays is bent or refracted when it 
passes through material. If the material 
through which the rays pass is completely 
heterogeneous in structure and orientation. 



Organization and Replication of DNA in Vivo 



267 



figure 20-3. X ray diffraction photographs of suitably hydrated fibers of DNA, showing the 

so-called B configuration. A. Pattern 
obtained using the sodium salt of 
DNA. B. Pattern obtained using the 
lithium salt of DNA. This pattern 
permits a most thorough analysis of 
DNA . ( Courtesy of Biophysics Re- 
search Unit. Medical Research Coun- 
cil. King's College. London.) 












♦ * ♦ < 





figure 20-4. The Watson-Crick dou- 
ble-stranded helix configuration of 
DNA. 



2(»S 



CHAPTER 20 



the emergent beam shows qo regularity when 

it is refracted. But it' the material is com- 
posed oi macromolecular units and or mo- 
lecular subunits spatially arranged in a reg- 
ular manner, then the emergent beam will 
form an X-ray diffraction pattern. This 
particular X-ray pattern can be used to 
identify units and subunits that arc repeated 
at regular intervals. It is known that each 
nucleotide in a DNA chain occupies a length 
of 3.4 A along the chain; this repetition is 
detectable by the characteristic X-ray dif- 
fraction pattern it produces — the black spots 
located symmetrically near the upper and 
lower edges of both photographs in Figure 
20-3. 

X-ray diffraction patterns have been ob- 
tained from the DNA of numerous species. 
In some cases the DNA was not removed 
from the nucleus; in other cases it was re- 
moved and also separated from the nucleo- 
protein protein. The spacings between DNA 
parts, and hence the X-ray diffraction pat- 
terns, depend upon the degree to which the 
DNA is hydrated. In all cases, provided 
that the DNA is similarly hydrated, essen- 
tially the same patterns attributable to DNA 
are found (Figure 20-3). In addition to 
the 3.4 A repetition, a study of these com- 
mon patterns shows other repeated units 
which can be explained only if DNA does 
not usually occur as a single strand. (On 
the other hand, X-ray diffraction studies 
show that chromosomal RNA usually is 
single-stranded. ) Here then is clear evi- 
dence for the existence of a secondary 
structure to DNA normally found in all 
chromosomes. (We might infer some sort 
of secondary organization for DNA from the 
independent observation of the equivalences 
A = T and G = C.) The simplest explana- 
tion consistent with the diffraction results 
of M. H. F. Wilkins and co-workers was 
proposed by J. D. Watson and F. H. C. 
Crick (1953a). They hypothesized that 
DNA is normally two-stranded (Figure 



20-4 ) ; each strand being a polynucleotide, 
and the two strands coiled around each other 
in such a manner that they cannot be sepa- 
rated unless the ends arc permitted to re- 
volve. This kind of coiling is called plecto- 
nemic (coiled like the strands of a rope) in 
contrast with paranemic coiling, which per- 
mits the separation of two coils without 
revolving their ends (just as two bedsprings 
pushed together can be separated). 

The Watson-Crick model for the second- 
ary organization of DNA macromolccules 
involves a double helix in which each strand 
is coiled right-handedly (clockwise). This 
coil direction is the same as that found in 
the secondary structure of amino acid chains. 
polypeptides. The model shows the pentose- 
phosphate backbone of each strand on the 
outside of the spiral (comprising the rib- 
bon), whereas the relatively flat organic 
bases projecting into the center (as bars) lie 
perpendicular to the long axis of the fiber 
(indicated by a vertical interrupted line). 
The backbone completes a turn each 34 A. 
Since each nucleotide occupies 3.4 A along 
the length of a strand, 10 nucleotides occur 
per complete turn, and each successive nu- 
cleotide turns 36° in the horizontal plane 
(so that 10 nucleotides complete the 360° 
required for a complete turn). 

The two helices are held together by 
chemical bonds between bases on different 
strands. The two strands can form a reg- 
ular double helix with diameter uniformly 
about 20 A only if the bases on different 
strands join in pairs, each pair composed 
of one pyrimidine and one purine. A pair 
of pyrimidines (being single rings) would 
be too short to bridge the gap between back- 
bones, whereas two purines (being double 
rings) would take up too much space. 
Moreover, it is found that the pyrimidine- 
purine pairing must be either between C and 
G or between T and A, for only in this way 
is the maximum number of stabilizing bond- 
ages between them produced. The type of 



Organization and Replication of DNA in Vivo 



269 



stabilizing bond holding the members of a 
base pair together is called a hydrogen bond 
or H bond. The base pairs, their H bonds 
indicated by interrupted lines, are shown in 
Figure 20-5; the hydrogens that are re- 
moved when the base pairs join the back- 
bones are included in the diagrams. The 
top half of the illustration shows the C-G 
(and G-C) arrangements. Note in the C-G 
pair that cytosine has been turned over (from 
left to right) relative to the way it was 
shown in Figure 19-3. Three H bonds are 
formed. Two occur between NFL. and O 
(the 6— NH, of C with the 6— O of G; 
the 2— O of C with the 2— NFF of G). 
One occurs between the 1 — N of C and the 
1 — NH of G. The GjC pair is identical to 
C;G shown except that, in this case, the 
base turned over is guanine. 

The bottom half of Figure 20-5 shows 
the other type of base pair (T:A or A:T, 
in which T and A have been turned over 
relative to the way they were shown in Fig- 
ures 19-3 and 19-4). In this pair only 
two H bonds are formed, one between the 
6 — O of T and the 6— NH,. of A; the other 
between the 1 — NH of T and the 1 — N of 
A. Although the hydrogen bond is a weak 
chemical bond as compared to the C — C 
bond, there are so many H bonds along a 
long double helix that the entire structure 
is fairly rigid and paracrystalline even when 
moderately hydrated. Note that the region 
surrounding two base-paired nucleosides 
can be separated into two portions relative 
to the pentoses. The smaller portion is 
called the minor groove (the region sur- 
rounding the lower parts of the base pairs 
shown in Figure 20-5), and the larger por- 
tion, the major groove (the region surround- 
ing the upper parts). 

Recall that the double helix configuration 
of DNA does not dictate the sequence of 
bases along the length of a chain. But re- 
member also that the sizes of the pyrimidines 
and purines and their H bonds do dictate 




Cytosine 



Guanine 



H 



--H-N 



/ 




Guanine 



Cytosine 



CH- 




O 



N-H 



\ i>— - u // \\ 



H 7 > 

Thymine 




Adenine 




O H 

Thymine 



FIGURE 20-5. 

Base pairs formed between single DNA strands. 



270 



( HAP I BR 20 



i k,i ki 20—6. ///<• opposite direction of the sugar-phosphate 
linkages in the two strands of a DNA double helix. 



5'CH 



H 2 C5' 




that A in one chain can pair only with T in 
the other chain — similarly C with G — to 
form a double helix of constant diameter 
whose strands are held together by the maxi- 
mum number of H bonds. Since A and T 
always go together (as do C and G), the 
equivalences A = T and C = G, derived 
from chemical analysis of DNA, become 
meaningful as the direct consequence of the 



secondary structure of DNA. In fact, these 
chemical equivalences provide the first in- 
dependent test of the Watson-Crick model 
constructed initially on the basis of the X-ray 
diffraction diagrams among other considera- 
tions. 

To form the maximum number of H bonds 
between a purine and a pyrimidine, it is 
necessary to represent one of the two as 



Organization and Replication of DNA in Vivo 



271 



being turned over, so that the number 1 
atoms of both face each other. This ar- 
rangement has an important consequence 
for the orientation of the two chains rela- 
tive to each other, as represented in the two- 
dimensional diagram, Figure 20-6. The 
bases in the chain at the right all face the 
accustomed way; those in the left chain are 
all turned over. For each base to join to 
its sugar in the same three-dimensional way, 
the sugars must be arranged as shown. No- 
tice, in proceeding downward from the top 
of the right chain, the P0 4 ~ linkages to sugar 
read 3'5', 3'5', and so on; reading down in 
the same way, however, the left chain is 
5'3', 5'3', et cetera, so that the member 
chains in a double helix run in opposite 
directions, as indicated by the arrows. 

The X-ray diffraction results, which led to 
the double helix hypothesis, do not tell us 
that all DNA in chromosomes is two- 
stranded, or that a double strand is never 
single-stranded at certain places or at cer- 
tain times. Such data prove only that, in 
the wide variety of organisms studied, a very 
appreciable part of the chromosomal DNA 
is not single-stranded. The base content 
and organization of DNA in viruses attack- 
ing bacteria have also been studied by chem- 
ical analysis and by X-ray diffraction. In 
the varieties T 2 and T 7 , for example, the 
data are entirely consistent with DNA's 
being present in the Watson-Crick double- 
helix configuration. In the mature bacterial 
virus particles of two other smaller varieties 
(called 0X174 and 0S13), however, the 
DNA is definitely single-stranded. This is 
reflected in the nonequivalence of A and T 
and C and G and the absence of those pat- 
terns indicating a secondary structure in the 
X-ray diffraction photographs. 

Whenever the DNA is in the double-helix 
configuration, we can consider one strand is 
the complement of the other, so that if the 
sequence of bases in one strand is known, 
the composition of the other strand can be 



determined. Thus, if one strand has the 
base sequence ATTCGAC, the other strand 
would have to contain TAAGCTG in the 
corresponding region. 

If DNA is genetic material, we expect 
DNA to be replicated just as accurately as 
genetic material. Since the base sequence 
in one strand is complementary to the se- 
quence in the other, we immediately see a 
simple way in which the double helix might 
be replicated: ] the two strands separate, 
and then each strand builds its complement. 
In this explanation, called the strand separa- 
tion hypothesis of DNA replication, each 
strand is visualized as a mold or template. 
We know that complex surfaces (like 
statues) can be copied exactly by making 
a mold which, in turn, can be used to make 
a second mold which is an exact copy of the 
original configuration. In the present case, 
the two complementary strands of DNA can 
be viewed as molds, or templates, for each 
other. One strand or both strands act as 
a mold on which the complementary strand 
is synthesized. Figure 20-7 shows one pos- 
sible sequence of events. At the top of 
this figure, the two strands are coming apart 
due to rupture of the H bonds. At the 
center the two single chains exist in the 
presence of single nucleotides or their pre- 
cursors. When the complementary free nu- 
cleotide approaches the single strand, its base 
is H-bonded. Then, after two or more nu- 
cleotides have bonded to the single strand, 
they are linked — perhaps by an enzyme — to 
start the new complementary strand. The 
bottom diagrams show sections of the com- 
plementary strands whose synthesis is al- 
ready completed. 

Experiments can be designed - to simul- 



1 Based upon the hypothesis of J. D. Watson and 

F. H. C. Crick (1953b, c). 

- Based upon those of M. Meselson and F W 

Stahl. 



272 



( II M'TF.R 20 



taneously test the hypotheses for both the 
double-helix structure of DNA and its repli- 
cation after strand separation. Remember 
that every pyrimidine or purine base nor- 
mally found in DNA contains two or four 
N atoms, respectively. Ordinarily, these are 
atoms of N-14, or light nitrogen. It should 
be possible to grow bacteria in a culture 



medium whose only nitrogen source is in the 
form o\' a heavier isotope, N-15, or heavy 
nitrogen. If so, after a number of genera- 
tions have passed, almost all of the DNA 
present will have been synthesized using 
heavy nitrogen. Suppose also that one can 
synchronize the multiplication of the bacteria 
containing heavy DNA. What will we ex- 








figure 20-7. Diagrammatic representation of the hypothesis of DNA 
replication after strand separation. 



Organization and Replication of DNA in Vivo 



273 



pect to happen if these bacteria are quickly 
washed, placed in a culture medium con- 
taining only light nitrogen, and permitted to 
continue their synchronous multiplication? 
The DNA should replicate each time the 
bacteria undergo cell division. During the 
first replication of DNA, the two strands 
containing heavy nitrogen should separate, 
and each should synthesize a complementary 
strand containing only light nitrogen. Thus, 
after one DNA replication, the density of 
the DNA molecules should be exactly mid- 
way between completely light and com- 
pletely heavy DNA. 

To test whether or not this expectation 
is actually observed, DNA is extracted from 
"all-heavy" bacteria and also from "all- 
light" bacteria. These extracts, serving as 
controls, first are ultracentrifuged separately 
and then together in a fluid medium con- 
taining cesium chloride. When a solution 
of cesium chloride is ultracentrifuged for 
about twenty hours, a gradient of densities 
is established because the concentration of 
cesium chloride is greatest at the bottom of 
the ultracentrifuge tube and least at the top. 
In the ultracentrifuge tube DNA assumes 
the position corresponding to its own density. 
In the density gradient the position of the 
DNA can be detected by its absorption of 
ultraviolet light at 2600 A. Two separate 
bands of DNA are found in the medium, one 
containing the all-heavy and the other the 
all-light DNA. When DNA is extracted at 
various time intervals after the originally all- 
heavy bacteria have been placed in the all- 
light nitrogen culture medium, the DNA 
band in the ultracentrifuge tube is observed 
to move from the all-heavy DNA position to 
a position exactly intermediate between the 
all-heavy and all-light positions (Figure 
20-8). This result is exactly what is ex- 
pected if after one replication the DNA is 
"hybrid'' in density. 

What would one expect to find after an 
additional DNA replication? In this case, 



the two strands of the hybrid DNA should 
separate, and light complementary strands 
should be made by both the light and heavy 
single strands. So, after a second replica- 
tion, half of the double-stranded DNA mole- 
cules should be all-light, and half should be 
intermediate between all-light and all-heavy 
(that is, they should be hybrid). In fact, 
the samples of DNA taken at later intervals 
show the single band at the intermediate 
position in the ultracentrifuge tube has be- 
come two bands, one at the hybrid position 
and one at the all-light position. It should 
be noted, moreover, that the time required 
for the change from all-heavy to all-hybrid 
molecules, or for the change from all-hybrid 
to half all-light and half hybrid molecules, 
is approximately the interval occupied by a 
bacterial generation. 

Although these results are consistent with 
the hypothesis of replication of double- 
stranded DNA following chain separation, 
they do not automatically exclude other pos- 
sible explanations. It might be claimed, for 
instance, that the double helix grows not 
by separation of strands followed by the 
synthesis of complementary ones, but by the 
addition of new double strand material to 
the ends of the original double strand. This 
alternative explanation can be tested in two 
ways. 

If the all-heavy molecules present initially 
grew by adding light material to their ends, 
they should be composed linearly of double 
strands that are successively heavy and light. 
It should then be possible for sonic vibra- 
tions to fragment the macromolecules into 
smaller segments, some all-heavy and others 
all-light. This result should be detectable 
in the ultracentrifuge tube by some of the 
sonicated hybrid DNA assuming the all- 
light and some the all-heavy positions. 
However, nothing happens; the DNA re- 
mains in essentially the same hybrid position 
whether or not it is sonically fragmented. 

A second test of the view that DNA syn- 



274 



CHAPTER 20 



a 



b 



EXP. NO. 



GENERATIONS 




0.3 



0.7 



1.0 



I.I 



1.5 



1.9 



2.5 



3.0 



figure 20-8. Test of the 
"replication after chain sep- 
aration" hypothesis, using 
the technique of density 
gradient centrif ligation. 

DNA was extracted from 
all-heavy (N-l 5-labeled) 
bacteria grown for different 
generation times on all-light (N-14-containing) medium. The extracts were subjected to ultra- 
centrifugation to position the DNA in the centrifuge tube according to its density. (Density 
increases to the right of the figure. ) DNA absorption of ultraviolet light is indicated by the 
bands in different photographs under a and the height of the peaks in the corresponding densi- 
tometer tracings under h. The rightmost band in the bottom two frames and the bund in the 
top frame represent all-heavy DNA. The leftmost band, seen clearly in all generation times 
after 1.5 generations, represents all-light DNA. The only other clear band is between the all- 
heavy and all-light ones. I his is the only band present after 1.0 generations, and represents 
DNA which is hybrid in density. Note that at 1.9 generations, half the DNA is all-light 
and half is hybrid in density (see row showing and 1 .9 mixed). {Courtesy of M. Mesel- 
son and F. W. Stalil, Troc. Nat. Acad. Sci., U.S., 44:675, 1958.) 



Oandl.9 
mixed 



Oand 4.1 
mixed 



Organization and Replication of DNA in Vivo 



275 



thesis is at the ends of double strands in- 
volves some of the following facts: When 
any sample of natural or native, double- 
stranded DNA is heated to an appropriate 
temperature (near 98° C), the H bonds are 
broken and the complementary strands sepa- 
rate. Double-stranded DNA's with high 

A 4- T 

— ratios become single-stranded at a 
G + C B 

lower temperature than do those with low 
ratios. This result is expected since high- 
ratio DNA is richer in A-T than low-ratio 
DNA, each pair of which has one less H 
bond than a C-G pair, so that less energy 
is needed to break the smaller total of H 
bonds. If the appropriately heated mixture 
is cooled quickly, the chains remain single, 
producing denatured DNA . That heat de- 
naturation followed by quick cooling pro- 
duces single strands from double helices can 
be confirmed by the loss of that part of the 
DNA X-ray diffraction pattern which de- 
notes polystrandedness. The change to 
single-strandedness is also accompanied by 
an increase of as much as 40% in the ab- 
sorption of ultraviolet light of 2600 A, so 
that single-stranded DNA is relatively hy- 
perchromic. It also is slightly denser than 
double-stranded DNA. If the hot mixture, 
containing denatured DNA, is cooled slowly, 
base pairing occurs and renatured DNA is 
obtained which shows a hypochromic effect 
and, from X-ray diffraction studies, evidence 
of double helices. 

The second test of endwise DNA synthesis 
involves converting double-stranded, all-light 
and all-heavy DNA to the single-stranded 
condition and locating the positions of the 
two types of single strands in the ultra- 
centrifuge tube. The "hybrid" double- 
stranded DNA is then made single-stranded 
and is ultracentrifuged. This preparation 
shows only two major components, one lo- 
cated at the all-light single-stranded position 
and the other at the all-heavy single-strand 
position. This result also is inconsistent 



with the hypothesis being tested. Not only 
do the two tests eliminate the view that ap- 
preciable endwise synthesis of DNA occurs 
in bacterial DNA, but they offer additional 
support for the hypothesis of replication 
after strand separation. 

Similar experiments yielding similar re- 
sults have been performed using the uni- 
cellular plant, Chlamydomonas, and higher- 
organisms, including man. The general 
agreement in the results of all these experi- 
ments apparently furnishes conclusive proof 
of the correctness of the Watson-Crick hy- 
potheses for the double-helix configuration 
of chromosomal DNA and for its replication 
after strand separation. 

Although the nuclear DNA of most or- 
ganisms is present as nucleoprotein, being 
combined with histones or protamines, the 
DNA in bacteria and in the viruses attack- 
ing them seems to exist uncombined with 
basic protein. • In the latter case, the DNA- 
containing fibers have a diameter of about 
25 A. Ordinary chromosomes are probably 
polynemic with respect to DNA double 
helices, 4 although the exact number of double 
helices per chromatid is not yet known with 
certainty. The basic fibril in protamine- 
containing sperm seems to be about 40 A 
in diameter, containing one DNA double 
helix plus basic protein. 3 In cells contain- 
ing histones, two DNA double helices bound 
side by side, plus the histone. form a fibril 
which is about 100 A thick.' 1 

So far, the evidence presents no clue as 
to how either end of a DNA polymer termi- 
nates. The possibility exists that DNA is 
a circular molecule, although this explana- 
tion still leaves the problem of how chain 
separation occurs if there are no free ends 
to revolve. When DNA is extracted from 
human sperm, about 0.1% of the "purified" 
material is reported to be composed of amino 



3 See H. Ris and B. L. Chandler (1964). 

4 See W. J. Peacock (1963). 



276 



( II M'TER 20 



acids. Digestion of the DNA at those 
points where it is joined to amino acid pro- 
duces extensive depolymerization. Such re- 
sults indicate that the amino acid sequence 

is short and. therefore, must sometimes be 
located internally rather than always being 
at the end of a DNA strand. On the aver- 
age, there is a sequenee of three amino acids 
per thousand nucleotides. The amino acid 
sequences appear bound to the phosphate of 
the DNA. not as a side ehain. but as an 
integral part of the molecule. Thus, the 
backbone of the DNA strand appears to be 
interrupted by short amino add sequences. 
About 33 c '( of the amino acids in DNA 
are the hydroxyl amino acids, serine and 
threonine, and 10 to 15% are glutamic acid. 
As expected, available evidence indicates 
that the amino acid attached to the phos- 
phate of DNA is often serine. The occur- 
rence of amino acids in the DNA of human 
leucocytes, calf thymus, and the vaccinia 
virus has also been reported. The DNA 

■"■See A. Bendich and H. S. Rosencranz (1963). 



in the larval salivary gland chromosomes of 
Drosophila also seems to be interrupted." 

If confirmed, these results are significant 
since they involve sequences of amino acids 
which arc so short that, when interstitial, 
they may be immune to the action of certain 
digestive enzymes. Such amino acid groups 
may be involved in: 

1. The bending of DNA double-helices 
(which are rather rigid) especially 
where DNA is much coiled on itself 

2. The mechanism by which strand sep- 
aration leading to replication occurs 

3. The separation of functional DNA 
units 

4. The functioning of individual DNA 
units 

5. The mechanism of crossing over 

6. Mutagenesis by agents capable of af- 
fecting amino acids. 7 

,; See reference to D. M. Steffensen (1963) on p. 

162. 

7 See 1. A. Rapoport and R. G. Kostyanovskii 

(1959). 



SUMMARY AND CONCLUSIONS 

DNA /'// vivo usually exists in the Watson-Crick double helix configuration and usually 
replicates, after the strands separate, by the formation oi complementary strands. 

In certain viruses, <2>X174 and <2>S13 for example, the DNA is single-stranded. In 
bacteria and bacterial viruses, the DNA exists uncombined with protein. 

In ordinary chromosomes, the following are found: a basic 40 A thick fibril con- 
taining protamine and one DNA double helix; a basic 100 A thick fibril composed of 
histone and two DNA double helices. A chromatid probably contains a number of 
100 A fibrils. 

The DNA molecule seems to be interrupted periodically by short amino acid 
sequences. 



Organization and Replication of DNA in Vivo 277 

REFERENCES 

Bendich, A., and Rosencranz, H. S., "Some Thoughts on the Double-Stranded Model of 
Deoxyribonucleic Acid," Progr. Nucleic Acid Res., 1 :219-230, 1963. 

Crick, F. H. C, "Nucleic Acids," Scient. Amer., 197:188-200. 1957. 

Jehle, H. n Ingerman, M. L., Shirven, R. M., Parke, W. C, and Salyers, A. A., "Replica- 
tion of Nucleic Acids," Proc. Nat. Acad. Sci., U.S., 50:738-746, 1963. 

Luzzati, V., "The Structure of DNA as Determined by X-ray Scattering Techniques," 
Progr. Nucleic Acid Res., 1:347-368, 1963. 

Meselson, M., and Stahl, F. W., "The Replication of DNA in Escherichia coli," Proc. 
Nat. Acad. Sci., U.S., 44:671-682, 1958. 

Peacock, W. J., "Chromosome Duplication and Structure as Determined by Auto- 
radiography," Proc. Nat. Acad. Sci., U.S., 49:793-801, 1963. 

Rapoport, I. A., and Kostyanovskii, R. G., "The Mutagenic Activity of Several Inhibitors 
of Cholinesterase," Doklady Akad. Nauk SSSR, 131:191-194, 1959 (in Russian). 

Ris, H., and Chandler, B. L., "The Ultrastructure of Genetic Systems in Prokaryotes 
and Eukaryotes," Cold Spring Harb. Sympos. Quant. Biol., 28:1-8, 1964. 

Sueoka, N., "Mitotic Replication of Deoxyribonucleic Acid in Chlamydomonas rein- 
hardi," Proc. Nat. Acad. Sci., U.S., 46:83-90, 1960. 

Watson, J. D., and Crick, F. H. C, "Molecular Structure of Nucleic Acids. A Structure 
for Deoxyribose Nucleic Acid," Nature, London, 171:737-738, 1953a. Reprinted 
in Classic Papers in Genetics, Peters, J. A. (Ed.), Englewood Cliffs, N.J.: Prentice- 
Hall, 1959, pp. 241-243. 

Watson, J. D., and Crick, F. H. C, "Genetical Implications of the Structure of Deoxy- 
ribonucleic Acid," Nature, London, 171:964-969, 1953b. Reprinted in Papers on 
Bacterial Genetics, Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, pp. 125- 
130. 

Watson, J. D., and Crick, F. H. C, "The Structure of DNA," Cold Spring Harb. 
Sympos. Quant. Biol., 18:123-131, 1953c. Reprinted in Papers on Bacterial 
Viruses, Stent, G. S. (Ed.), Boston: Little, Brown, 1960, pp. 193-208. 

See bibliography and all but last portion of Supplement IV. 

QUESTIONS FOR DISCUSSION 

20.1. Can you draw any conclusions from the observation that most of the multi- 
cellular organisms studied are richer in A + T than C -f- G? 

20.2. Among the DNA molecules contained in a genome, why is it expected that 
many would differ in base sequence and content? 

20.3. How many different base pairs normally occur in a double helix of DNA? 
What are they? 

20.4. If a coil is right-handed when looked at from one end, is it also right-handed 
when seen from the other end? 

20.5. What would you have expected to see in the ultracentrifuge tube following sonic 
treatment of DNA or following the conversion of DNA to its single-stranded 
condition, if synthesis had occurred at the ends of the double DNA helix? 

20.6. What evidence can you give that heating double-helix DNA causes the strands 
to separate? 



278 CHAPTER 20 

20.7. When is DNA single-stranded? 

2o. s. A double helix oi DNA lias a base sequence \l I AG< A on one strand. Can 
you complete an inversion after breaking the backbone at two places on this 
single strand? Explain. Can you complete an inversion if the backbone of the 
complementary chain is also broken at exactly the same two levels? Explain. 

20.9. Given two double helices whose backbones are broken at the places indicated 
by periods: 

ATCG.GCAT AT.TAG 
TAGC.CGTA TA.ATC 

draw the base sequences which can occur following reciprocal translocation 

between double helices. 

20.10. In what respects is Figure 20-7 incorrect? 

20.11. M. Green and M. Piha (1963) report that the G + C ratio of a number of 
human and animal carcinogenic viruses is similar and lower than that of com- 
parable nontumorigenic viruses. Discuss the possible implications of this anomaly 
for the origin and action of tumor-inducing viruses. 

20.12. Locate on Figure 20-6 the minor and major grooves of the two pairs of nucleo- 
tides shown. Discuss the accuracy of this figure. 



NH, 




Chapter 21 

REPLICATION OF DNA 
IN VITRO 




o o o 

O— P— O— P— o— p— o- 

o- o- o- 



E 



Iarly in this book (p. 10) we 
assumed self-replication to be 
a characteristic of the genetic 
material. In light of the indirect evidence 
that chromosomal DNA is genetic material, 
it is of great interest to learn as much as 
possible about the replication of the DNA 
double helix. Although the evidence (Chap- 
ter 20) is fairly conclusive that comple- 
mentary chains are synthesized after chain 
separation, no evidence has yet been pre- 
sented about how this replication is accom- 
plished. Figure 20-7 and the discussion on 
page 271 only postulate a mechanism which 
includes an enzyme that joins the nucleotides 
forming a new complementary strand. 

Since the linear combination of nucleo- 
tides undoubtedly requires energy, consider 
the possible source of this energy. Consid- 
erable chemical energy is contained in the 
ribotide, adenosine triphosphate (ATP), a 
riboside 5 '-triphosphate (Figure 21-1). 
The energy hitherto needed to bond two 
phosphates to adenosine monophosphate be- 
comes available when ATP reacts with other 
nucleotides or acids and loses its two ter- 
minal phosphates as inorganic pyrophos- 
phate. Because ATP is known to supply 
the energy for many chemical reactions in 
the cell, it is reasonable to suppose that it 
may also supply the energy needed to join 
individual deoxyribotides to a DNA strand 
during replication. 

Since DNA removed from the nucleus and 
separated from protein still retains what ap- 
279 



OH OH 

FIGURE 21-1. 

Adenosine 5' -triphosphate (ATP) (APPP). 

pear to be its main characteristics in situ 
(in the living cell), we may well be able to 
study DNA synthesis under nonliving condi- 
tions. What should we extract from cells 
in order to study DNA synthesis in vitro? 
Basically, we ought to use all the apparatus 
the cell normally utilizes for this function. 
From the strand separation viewpoint, DNA 
is needed to serve as a template for DNA 
synthesis, so the extract should contain the 
cell's DNA. ATP is added to the extract 
as the source of energy required for the 
synthesis. MgCl. can also be added; since 
the magnesium ion, Mg++, is known to 
activate many enzymes, perhaps it will also 
act on the one required for DNA strand 
formation. 

How can we tell whether DNA is syn- 
thesized in the extract? Any crude cellular 
extract would be expected to contain DNases. 
These enzymes might depolymerize or other- 
wise degrade DNA as fast as — or faster 
than — any process synthesizing DNA. The 
problem of identifying DNA synthesis in the 
absence of a net increase in DNA quantity 
can be solved by preparing the deoxyribo- 
side thymidine with radioactive C u incor- 
porated in its pyrimidine and adding this 
•'hof chemical to the extract. If any radio- 
actively-labeled thymidine is incorporated 
into DNA, it would happen as part of the 
synthetic reaction, since incorporation into 
DNA only occurs during synthesis. 



280 



CHAPTER 21 



Finally, we ought to obtain the extract 
From cells that are growing and dividing 
rapidly, for these cells are likely to contain 
the greatest amount of functional apparatus 
for DNA synthesis. In line with this rea- 
soning, an experiment is performed with an 
extract of the bacterium Escherichia coli? 
ATP, Mg " ions, and radioactive thymidine 
are added and the pH is adjusted to suit 
experimental conditions. After an incuba- 
tion interval (about 30 minutes), the pH 
is made suitably acidic for precipitating a 
DNA polymer; single deoxyribosides — that 
is, monomers — remain soluble. The acid 
precipitate is washed many times until it is 
certain that the DNA precipitate is not 
contaminated by adsorbed deoxyribosides. 
When the DNA is examined, it is found to 
be only slightly radioactive (50 counts per 
unit time as compared with 5 million counts 
in the thymidine substrate added). In fact, 
so little thymidine is incorporated in the 
DNA that it is 10,000 times too small to 
be detected by ordinary chemical analysis. 
Nevertheless, the radioactivity is without 
doubt due to thymidine incorporated into 
DNA and can be released from the precipi- 
tated DNA by treatment with DNase. 

Although this result is not quantitatively 
impressive, the process furnishes C ,4 -thymi- 
dine-labeled, acid-precipitable, DNase-sensi- 
tive DNA as the end product. The amount 
of this labeled material formed can be used 
to determine the effect of changes in the ex- 
perimental procedure. This fact has already 
led to a change in the procedure and to a 
better understanding of the nature of the 
reaction. 

Reactions that produce derivatives of 
adenosine commonly start with ATP as one 
of the reactants. Similarly, derivatives of 
uridine, cytidine, and guanosine involve their 
respective triphosphates and the liberation 

1 The preceding and following account is based 
primarily upon work by A. Kornberg and his 
associates. 



oi inorganic pyrophosphate. Such facts lead 
to the conclusion that the fundamental unit 
in the formation of diribotides or polyribo- 
tides is the riboside 5'-phosphate, activated 
in the form of riboside 5 '-triphosphate. It 
is reasonable, therefore, to assume that the 
active building block of polydeoxyribotides 
is the deoxyriboside 5'-triphosphate. 

If this molecule is the building block, the 
ATP added in the in vitro experiments may 
be converting various deoxyribosides — al- 
ready present or added to the extract — to 
the 5 '-triphosphate condition (making, for 
instance, C 1 '-thymidine 5'-triphosphate). 
This view is supported since DNA synthesis 
occurs in vitro when labeled thymidine 5'- 
triphosphate (T*PPP) is used instead of 
labeled thymidine (T*)+ATP (APPP). 

To learn more about the ingredients es- 
sential to DNA synthesis, the initial extract, 
obtained from the sonic treatment of bac- 
teria, is fractionated and its protein, concen- 
trated. This procedure results in a nearly 
4,000-fold increase in synthetic activity. 
From this and other evidence, it becomes 
clear that the presence of a protein catalyst — 
the enzyme E. coli DNA polymerase (or 
DNA duplicate) — is essential for the syn- 
thetic reaction to take place. 

Once E. coli DNA polymerase is concen- 
trated, it is possible to obtain a large net 
increase in DNA (final amount minus initial 
amount). Such a net increase, however, is 
obtained only if the 5'-triphosphates of all 
four deoxyribosides commonly found in 
DNA are added to the incubation mixture. 
Deoxyriboside 5'<7/phosphates are not active, 
nor are riboside 5'-triphosphates. The other 
requirements for net increase in DNA 
amount are: 

1. The presence of already-formed DNA 
molecules of high molecular weight 

2. Mg+ f ions 

3. DNA polymerase. 

The already-formed, high molecular weight 



Replication of DNA in Vitro 



281 




P-P--P 



* 



>DNA 




>DNA 



figure 21-2. Growth of a primer DNA strand at its nucleoside end (left) and nucleo- 
tide end (right). Arrows show 5' positions of subsequent degradation by micrococcal 
DNase plus splenic phosphodiesterase. 



DNA may come from a plant, animal, bac- 
terium, or virus. Similar DNA-synthesizing 
extracts can be prepared from other bacteria, 
calf thymus, and various animal tissues. 

Extended and Limited Syntheses 

Using E. coli preparations, we can obtain 
an extended synthesis of DNA which pro- 
duces twenty or more times as much DNA 
as was initially present. In this case, there- 
fore, 95% or more of the DNA present at 
the end must have been synthesized from 
the triphosphates added as substrate, the ex- 
tended synthetic reaction proceeding until 
the supply of one of the four triphosphates 
is exhausted. One inorganic pyrophosphate 
is released for each deoxyribotide incorpo- 
rated into DNA. 

Although extensive synthesis of DNA does 
not occur if only one of the deoxyriboside 
5'-triphosphates is added as substrate, some 
incorporation of this nucleotide into the 



DNA strand occurs in what is called a lim- 
ited reaction. By what mechanism does the 
nucleotide add on to the pre-existing DNA 
strand? In this case, the already-present 
DNA must provide a point of linear attach- 
ment for newly-forming DNA, thereby func- 
tioning as a primer. Suppose that the only 
triphosphate added to the substrate is de- 
oxycytidine 5'-triphosphate whose innermost 
phosphate carries radioactive P A - (dCP*PP). 
The two possible ways in which the DNA 
strand might lengthen are shown at the left 
and right of Figure 21-2; the DNA strand, 
present as primer, is shown enclosed in 
brackets. The primer strand can be con- 
sidered to have a nucleotide end (top of the 
figure) — to which pyrophosphate, P-P, is 
added in the diagram to the right — and a 
free 3'-OH nucleoside end (bottom of illus- 
tration). (The removal of a sugar and base 
by a single break at the 5' position involves 
the removal of a nucleoside at the nucleoside 



282 



CHAPTF.R 21 



end and the removal oi a oucleotide at the 
nucleotide end. i The diagram at the left of 
the figure shows the </('P adding on to the 
nucleoside end by the formation o\ a 3' 
linkage between P* and the sugar at the end 
of the chain. P-P being split oil the </CP*PP. 
The diagram at the right shows </CP*PP 
adding on to the nucleotide end o\ the chain 
b\ linkage to the 5' position of the end nu- 
cleotide which supplies the pyrophosphate 
that splits off. In brief, the DNA strand 
might be lengthened by the addition of a 
nucleotide at either the 3' position of the 
nucleoside end or at the 5' position of the 
nucleotide end. 

It is possible to distinguish between these 
two alternatives by treating the product of 
a limited reaction first with DNase from 
micrococci to enhance the action of another 
added enzyme, splenic phosphodiesterase. 
The latter enzyme degrades DNA by break- 
ing the strand at all the 5' positions, so that 
deoxyriboside 3'-monophosphates are pro- 
duced. This position of breakage is indi- 
cated by the arrows in Figure 21-2. If the 
strand lengthens according to the diagram at 
the right of the figure, radioactive P 32 should 
be found in phosphate attached to deoxycy- 
tidine. and P* should not be part of the 3'- 
deoxyribotides of A. T. or G. If. on the 
other hand, attachment is to the 3' position 
at the nucleoside end of the strand, then, as 
can be seen in the diagram at the left of the 
figure, P* should not occur in inorganic 
phosphates but should sometimes appear in 
other deoxyriboside 3'-monophosphates be- 
sides those containing C. The experiment 
gives the latter result; that is, not only is 
P* frequently present in all four kinds of 
deoxyriboside 3'-monophosphates. but it is 
absent from inorganic phosphate. 

An additional test of whether the DNA 
strand grows at its 3' position involves treat- 
ing the limited product with a different en- 
zyme, snake venom diesterase. This en- 
zyme digests DNA by breaking the bond 




^DNA 



P-P--P 



FIGURE 21-3. Degradation of a primer DNA 
strand, which has grown at its nucleoside end, 
by snake venom diesterase. The 3' positions 
of degradation (arrows) occur sequentially 
starting at the nucleoside end. 



between the phosphate and sugar at the 3' 
position, starting at the nucleoside end of 
the strand and proceeding toward the nucleo- 
tide end. Thus, the DNA is gradually di- 
gested into deoxyriboside 5'-phosphates, as 
indicated by the arrows in Figure 21-3. 
When the limited product is treated this way, 
the result — as expected — is that almost all 
the radioactivity has been removed from the 
strand even though only a minute portion 
of the DNA has been digested. Other re- 
sults clearly show that the product of a lim- 
ited reaction is DNA with one or very few 
deoxyribotides added to the nucleoside end 
of the strand. Still other evidence supports 
the view that the 3' point of lengthwise link- 



Replication of DNA in Vitro 



283 



age is the same when net DNA is greatly 
increased as it is when the limited reaction 
occurs. 

Characteristics of Synthesized DNA 

That a DNA primer strand can be length- 
ened in vitro, no matter which of the four 
common deoxyribosides happens to be at 
the nucleoside terminus, is consistent with 
the independence of DNA primary structure 
upon base sequence. Consider the evidence 
that the DNA synthesized in vitro has the 
characteristics of DNA synthesized in vivo. 
The physical characteristics of DNA samples 
consisting of 90% or more of the product 
synthesized in vitro are similar to those of 
DNA isolated from calf thymus insofar as 
sedimentation rate and viscosity are con- 
cerned. Such results indicate that the prod- 
uct has a high molecular weight (about six 
million) and, usually, is not single-stranded. 
In support of the latter inference is the find- 
ing that the macromolecular structure of the 
in vitro product is destroyed when heated 
for 10 minutes at 100° C, an expected result 
if this treatment is to produce single strands 
— denatured DNA which collapses to form 
compact, randomly-coiled structures. Like 
thymus DNA, the enzymatic product shows 
the same type of increase in ultraviolet ab- 
sorption following digestion with pancreatic 
DNase. 

If the synthesis in vitro occurs as it does 
in vivo, we might expect single-stranded 
DNA to serve in in vitro synthesis as well 
as, or better than, double-stranded DNA. 
In fact, the single-stranded DNA, isolated 
from the virus </>X174, is excellent for this 
purpose, and heat-treated DNA is better 
than unheated DNA. Moreover, the prep- 
arations containing the most active DNA 
polymerase do not work well with double- 
stranded DNA unless it is first heated or 
treated with DNase. The DNA produced 
in extended syntheses behaves as though it 
is primarily two-stranded, but differs from 



native double-stranded DNA by appearing 
to be markedly branched in electron micro- 
graphs and by being readily renatured after 
heat or alkaline denaturation. These results 
suggest that strand separation is usually in- 
complete in the in vitro system. 

In view of these results, we can conclude 
that the physical characteristics of the DNA 
synthesized in vitro and in vivo are extremely 
similar, though not identical. Synthesis 
clearly involves single strands which pro- 
duce double strands probably held together 
by H bonds. 

One can also study the detailed chemical 
and physico-chemical characteristics of the 
in vitro synthesis of DNA. If single strands 
produce double strands by forming comple- 
mentary structures, then the capacity to form 
a complementary strand should depend upon 
the presence in the substrate of purine and 
pyrimidine bases which can form appropriate 
H bonds with the bases in the added DNA. 
In other words, in an extensive synthesis, 
pre-existing DNA should serve as a template 
for the synthesis of complementary DNA 
strands. Figure 21-4 shows some pyrimi- 
dine and purine bases which do not naturally 
or frequently occur in DNA as well as the 
four principal types which do. The un- 
natural or infrequent bases include: uracil 
and 5-bromo uracil (both of which are ex- 
pected to have the same H-bonding capaci- 
ties as thymine); 5-methyl cytosine and 5- 
bromo cytosine (both of which are expected 
to have the same H-bonding capacities as 
cytosine); and hypoxanthine (which has two 
of the three H-bonding sites found in gua- 
nine). If A in the single-strand preformed 
DNA dictates its complement — by specify- 
ing that the complementary base is a pvri mi- 
dine that provides the proper sites for H- 
bonding A — then one would expect thai 
uracil or 5-bromo (or 5-fluoro) uracil can 
substitute for the thymine in thymidine 5'- 
triphosphate. 

When the substrate used contains do- 



284 



CHAPTER 21 




CH; 



Thymine 




Uracil 




5-Bromo uracil 



NH 




NH, 




CH; 



CT N' 
H 

5-Methyl cytosine 



NH. 





^HN 

NH^ NT N 
H 
Guanine 



HN 




Hypoxanthine 



figure 21-4. Various bases utilized 
in in vitro synthesis of DNA. Ar- 
rows point to groups capable of 
typical H-bonding. 




oxyuridinc 5'-triphosphate (or 5-bromo 
or 5-fluoro deoxyuridine 5'-triphosphate), 
</CPPP, </APPP, and </GPPP, DNA syn- 
thesis occurs. Similarly, 5-mcthyl cytosine 
or 5-bromo or 5-fluoro cytosine can substi- 
tute for cytosine. On the other hand, the 
substitution of hypoxanthine for guanine 
does not support DNA synthesis as well as 
do the other substitutions mentioned. This 



result is expected from the hypothesis under 
test, since the former has one less H-bonding 
site than the latter. Moreover — as pre- 
dicted — uracil, 5-bromo uracil, and 5-fluoro 
uracil will not substitute for C, A, or G in 
dCPPP, <7APPP, or JGPPP, respectively. 
Similarly, 5-mcthyl, 5-bromo, and 5-fluoro 
cytosine replace cytosine specifically; hypo- 
xanthine replaces only guanine. Although hy- 



Replication of DNA in Vitro 



285 



poxanthine has the same H-bonding groups 
as thymine, it does not replace thymine, 
probably because the A-hypoxanthine pair, 
being composed of two purines, takes up too 
much space to fit the regular double-helix 
configuration. These results support the 
hypothesis that normally the /'// vitro syn- 
thesis of DNA is dependent upon the forma- 
tion of complementary purine-pyrimidine 
pairs — A with T and C with G — just as is 
the case in vivo. 

The DNA synthesized in vitro from the 
usual four deoxyriboside 5'-triphosphates 
can be analyzed chemically. The analysis 
shows that in the in v/fro-synthesized DNA, 
A equals T and C equals G, just as they do 
in natural DNA, even though the relative 
concentration of the four triphosphates in 
the substrate is widely distorted. Not only 
do total pyrimidines equal total purines in 
"synthetic'" DNA, as described — whether a 
moderate or a large amount is synthesized — 
A + T 



but the particular 



ratio of the primer- 



G + C 

template is reproduced faithfully in the syn- 
thesized product (see Figure 8 in Supple- 
ment V). In this respect, the product is a 
replica of the primer-template, as is strik- 
ingly illustrated by the following experiment. 
After a long period of incubation of all the 
usual components except pre-existing DNA, 
a linear deoxyribotide polymer composed 
only of A and T is formed spontaneously. 
When this polymer is added as the pre-exist- 
ing DNA, even though all four usual triphos- 
phates are present in the substrate, the result 
is an extensive synthesis of material which 
contains A and T only, with no trace of C 
and G (again see Figure 8 in Supplement V). 
It has been mentioned that for each nu- 
cleotide added to the end of the DNA strand, 
an inorganic pyrophosphate, PP. is liberated. 
When PP is added to the usual synthesizing 
complex in great excess (about one hundred 
times the concentration of the triphos- 
phates), the synthetic reaction is inhibited 



by about 50%. This observation implies 
the reversibility of DNA synthesis in vitro. 
It was also mentioned earlier that in a 
limited reaction, </CP*(PP) can add onto 
a strand terminating in all four types of 
nucleotides (dCP, TP, d AP, and </GP). 
However, this statement does not suggest 
that dCP* joins linearly to each nucleotide 
with equal frequency, or that any nucleotide 
joins to all others with equal frequency. 
Other results indicate that the limited re- 
action does not add nucleotides to the end 
at random; this reaction probably involves 
the repair of the shorter strand of a double 
helix, the particular nucleotide added being 
specified in the usual way by the bases pres- 
ent in the longer strand. In other words, 
the shorter strand acts as a primer and the 
longer strand as a template. 

Dinucleotide Sequences 

What is the linear arrangement of nucleo- 
tides in the DNA synthesized in an extended 
reaction? If it is genetic material, different 
linear segments of DNA can represent dif- 
ferent genes, and the differences among 
genes would lie in the sequence of their 
organic bases. Considering only the four 
usual deoxyribotides, how many different 
sequences of two nucleotides are possible? 
The first nucleotide can be one of four, and 
so can the second, making a possible 4 times 
4. or 16 different linear arrangements in di- 
nucleotides. The orders in dinucleotides 
can be determined experimentally as follows: 
one of the four triphosphates added as sub- 
strate is labeled with P 32 in the innermost 
phosphate, the other three are not. Ex- 
tended synthesis is permitted during which 
the P* attaches to the 3' of the sugar of the 
nucleotide which is its linear neighbor (refer 
to the left part of Figure 21-2 and to Figure 
9 in Supplement V). This linear neighbor 
can be identified by digesting the synthesized 
product with micrococcal DNase and splenic 
phosphodiesterase. Recall that the latter 



286 



CHAP II R 21 



produces deoxj riboside 3'-monophosphates 

In breaking the chain at 5'. Consequently, 
the labeled phosphate ( P* ) is found joined 
at the }' position of the deoxyriboside just 
anterior to the one with which it entered 
the ON. A strand. The digest is then an- 
alyzed to see how frequentl) P* is part of 
</ A y-P*. T 3'-P*, dC 3'-P*, and dG 3'-P*. 
If the P* were originally in </AP*PP. we 
would then know the relative linear fre- 
quencies of TA, AA, CA. and GA. By 
carrying out this procedure three more times, 
labeling a different one of the triphosphates 
each time, the relative frequency of all six- 
teen sequences can be determined. 

Such nearest-neighbor analyses have been 
made of the DNAs synthesized using a num- 
ber of different preformed DNAs. As al- 
ready mentioned, the DNA isolated from 
particles of <£X174 is single-stranded; chem- 
ical analysis reveals its base content to be 
A = .246; T = .328; C = .185; and G = .242. 
It synthesis requires the formation of com- 
plementary base pairs, an extended reaction 
carried out with <£X174 DNA and all four 
triphosphates labeled and stopped after 20% 
synthesis has occurred, should produce com- 
plementary, labeled DNA composed of 
.328 A; .246 T; .242 C; and .185 G. The 
values found are exactly those expected. If 
both old and new strands are used as tem- 
plates, one expects and finds after a 600% 
synthesis that A = T and C = G. More- 
over. A and T are expected to have fre- 
quencies of M.328 -f .246), or .287, with 
C and G = .224. Again experimentation 
confirms expectation. 

One can perform the 20% synthesis on 
four occasions, each time labeling a different 
triphosphate. In this way, nearest-neighbor 
analyses are made, and the relative frequen- 
cies of all sixteen dinucleotide sequences de- 
termined. All sequences are found and in- 
clude, for instance. .054 GA; .064 TC; .052 
CT; and .069 AG. Nearest-neighbor anal- 
yses can also be made after 600% synthesis. 



II the complementary strand is synthesized 
in the same direction as the template strand, 
from the results of the 20 r ; syntheses one 
expects that GA CT ' .(.054 + .052 ) 

.053; and TC AG = .067. If. on the 
other hand, the two complementary strands 
are synthesized in opposite directions, the 
expectations are GA = TC ' •_. (.054 + 
.064) = .059; and CT = AG = .061. The 
values observed (.058 GA; .065 TC; .064 
CT; and .058 AG) clearly follow those ex- 
pected for complementary strands synthe- 
sized in vitro in opposite directions, as do 
the values obtained for the other dinucleo- 
tide sequences. 

Chemical analysis of Mycobacterium phlei 
DNA yields .162 A; .165 T; .335 C; and 
.338 G. If extended syntheses permitting 
nearest-neighbor analyses are performed, one 
can determine the relative amount of base X 
incorporated into DNA from the sum of the 
separate frequencies with which XA. XT, 
XC, and XG occur. When X is, in turn, 
A, T. C, and G, the relative frequencies are 
found to be .162, .164, .337. .337. respec- 
tively. Thus, what is already demonstrated 
via chemical analyses is independently 
proved via nearest-neighbor analysis — 
namely, that the product of an extended 
synthesis has the same base frequencies as 
the natural two-stranded DNA used as 
primer-template. 

The question of whether or not the se- 
quences of bases along a strand is random 2 
can be decided by using calf thymus DNA 
as primer-template and determining all the 
dinucleotide frequencies in the product. 
Among the frequencies observed, CG is .016 
and GC. .044. Had these two dinucleo- 
tides occurred with equal frequency, the 
hypothesis of a random nucleotide sequence 
in a strand would have been supported. 
Since the frequencies are clearly different, 
however, base sequence in a strand is not 

-' In this connection see also J. H. Spencer and 
E. Chargaff (1963). 



Replication of DNA in Vitro 



287 



random. The nonrandomness of base se- 
quences is also supported by experimental 
results which reveal that: 70% of the bases 
are distributed so that three or more pyrim- 
idines (and hence purines) occur in succes- 
sive linear nucleotides; linear sequences of 
five successive T's exist; and 4.9% of a 
given DNA contains sequences of eight or 
more pyrimidines in succession. 

Is the nearest-neighbor frequency the same 
when native DNA is used as primer-template 
as when DNA synthesized from this native 
DNA is used? With synthesized calf thy- 
mus DNA as primer-template, the nearest- 
neighbor frequencies of the newly synthe- 
sized product are essentially the same (for 
example, CG is .011 while GC is .042) as 
they are in the product formed using calf 
thymus DNA as primer-template (above). 
Consequently, as revealed by nearest-neigh- 
bor analysis, the products of synthesis are 
identical when native DNA and when DNA 
synthesized from native DNA serve as 
primer-template. 

A + T 
The — — L — ratio is 1.25 for calf thymus 
C + G y 

DNA and 1.29 for B. subtilis. Even though 
these base ratios are very similar, it is un- 
likely that one of these is a molecular poly- 
ploid of the other (see p. 265 ). In fact, the 
dinucleotide sequences determined for DNA 
synthesized from the bacterial DNA (for ex- 
ample, CG is .050 and GC, .061) are quite 
different from those of calf thymus (.016 
and .044, respectively). Other work gen- 
erally shows that in higher plants and ani- 
mals CG is lower than expected on a random 
nucleotide sequence, whereas the reverse 
occurs in bacteria. It should also be noted 
that different normal or neoplastic tissues of 
the same individual give nucleotide neighbor 
frequencies which are not demonstrably dif- 
ferent. We conclude, therefore, that each 
type of natural DNA has unique and repro- 
ducible dinucleotide sequences, not pre- 
dictable from its base composition. 



De Novo Synthesis of DNA 

The double-stranded polymer of A and T 
appearing de novo, referred to earlier, can 
be used as primer-template to study its di- 
nucleotide sequences. Only the AT and TA 
sequences are found; thus, it appears that 
A and T occur in perfect alternation in a 
strand, forming what is called a copolymer 
of AT, or dAT(d-AT). The de novo syn- 
thesis of a dAT, sufficiently long to serve 
as template for extended synthesis, requires 
a lag period of a few hours, during which 
</APPP, TPPP, Mg++, and E. coli DNA 
polymerase are incubated together. It has 
been hypothesized that during the lag period, 
the E. coli DNA polymerase catalyzes the 
de novo formation of single strands from 
single deoxyriboside 5'-triphosphates. Once 
the strand is started, the short strand or 
oligodeoxyribotide can serve as a primer to 
lengthen itself. In the presence of d/KPPP 
or TPPP alone, dAT adds one or two units 
of <r/AP or TP per chain in a limited reac- 
tion. Neither dCP nor dGP is incorporated 
into dAT when dCPPP and or dGPPP are 
added to the substrate, whether or not 
JAPPP and TPPP are also present. It is 
clear, therefore, that both limited and ex- 
tended synthesis of dAT require base-pair- 
ing. It is possible that a limited reaction 
involves a single strand of dAT which is 
folded so as to base-pair with itself except 
at the nucleotide end. So, once the dAT 
strand is long enough, the polymerase can 
use it as a template for base-pairing syn- 
thesis. Note that no lag period occurs in 
limited or extended base-pairing syntheses. 
In the absence of pre-existing DNA and 
alter a lag period. E. coli polymerase — in 
the presence of Mg+ + and high concentra- 
tions of dCPPP and dGPPP — catalyzes the 
de novo formation of another double- 
stranded polymer containing only C and G. 
Nearest-neighbor analysis shows only two 
dinucleotide sequences, CC and GG. 
Clearly, this polymer, called dGdC (or 



2SS 



( hap iik 21 



d-GC), is composed of two homopolymers; 
one strand contains onlj C's and the other 
onlj G's, the two strands base-pairing 
to form a double helix. Whereas A = T 
after extensive synthesis of dAT, most 
products of dCklC synthesis show 56 to 
81% dGP. 

Both dAT and dGdC illustrate that the 
base sequence in certain strands is not ran- 
dom. That such strands can have biological 
significance is strikingly supported by the 
discover} of a "natural dAT" polymer in 
the sperm of a certain crab. Nearest- 
neighbor analysis shows that this polymer, 
which comprises about 30% of the total 
DNA content, contains A and T in strict 
alternation in 93% of the dinucleotide se- 
quences. About 3% of the bases associated 
with this natural dAT are G or C, and all six- 
teen dinucleotide sequences are found. One 
might suspect that the G- and C-containing 
nucleotides are contaminants of the typical 
DNA comprising 70% of the total. If they 
are, extensive in vitro synthesis using natural 
dAT as primer-template should occur just as 
rapidly whether or not the substrate of 
clAPPP and TPPP has </CPPP and </GPPP 
added to it. One finds, however, that the 
replication rate in the absence of the latter 
two triphosphates is only 19% of that ob- 
tained in their presence, supporting the view 
that G and C bases are an integral part of 
the "natural dAT" polymer. 

Calf Thymus DNA Polymerase 

All in vitro DNA syntheses so far discussed, 
whether they occur de novo (by an un primed 
initiation of single strands) or involve com- 
plementary base-pairing (as in limited and 
extensive syntheses ). require the presence of 
E. coli DNA polymerase. The DNA poly- 
merase isolated from calf thymus cannot 
form dAT or dGdC de novo; in other words, 
it is apparently incapable of unprimed DNA 



synthesis. 1 Some evidence has been ob- 
tained that M-bond, base-pairing, template 
synthesis by calf thymus DNA polymerase 
begins once chain length exceeds twenty 
monomers. The DNA polymerase from calf 
thymus, like that from E. coli, uses DNA 
as a template to make DNA which resembles 
native DNA in primary and secondary struc- 
ture, composition, sequence, and molecular 
size; the synthesized DNA also does not 
completely undergo strand separation upon 
heat denaturation. 

Biological Replication in Vitro 

Insofar as extensive synthesis of DNA in 
vitro is concerned, all, or almost all the 
physical and chemical characteristics of the 
product are consistent with the view that 
this is a biological process. However, there 
are several differences between the in vitro 
and the in vivo processes. Apparently E. 
coli DNA polymerase works more slowly 
in vitro than in vivo. Calf thymus DNA 
polymerase has a maximum yield of 100%. 
Both enzymes make DNA which does not 
strand separate completely when heat de- 
natured; this behavior may be correlated with 
the branched DNA made in vitro by the E. 
coli enzyme. All these differences may re- 
sult from contaminants in vitro which are 
not present in vivo, and from changes which 
occur in isolating the in vivo DNA for in 
vitro syntheses. When these possibilities are 
considered, there is little doubt that exten- 
sive synthesis is performed in vitro in essen- 
tially the same manner as in the living cell 
and that it produces essentially the same 
product. 

Finally, special attention should be given 
to the DNA polymerases essential for the 
extended biological synthesis of DNA. Pre- 
viously-known enzymes are specific in that 
they act upon one or a few particular sub- 
strates usually modified in the same way. 



By N. Sueoka. 



•See F. J. Bollum (1963, 1964). 



Replication of DNA in Vitro 



289 



For example, trypsin breaks peptide bonds 
only at places in a polypeptide chain where 
lysine or arginine arc present. DNA poly- 
merases are unique in that they take direc- 
tions from a long template; the strand act- 
ing as template dictates which particular 
monomers can be added by DNA polymerase 
to the strand acting as primer. It should be 
noted that, in vitro, ribonucleotides can be 
incorporated into terminal positions in 
DNA ' and that, using DNA as a primer- 
template, E. coli DNA polymerase can use 
a mixture of riboside and deoxyriboside 
triphosphates to synthesize complementary 
strands that contain both ribo- and deoxy- 
ribotides.' ; No evidence exists, however, 
that such mixed RNA-DNA strands have 
biological or genetic significance. 

DNA Synthesis from RNA 

Using poly {A + U), the double helix com- 
posed of the homopolymer of adenylic acid 
base-paired with the homopolymer of uri- 
dylic acid, E. coli DNA polymerase activated 
by MgCL can be tested ' for ability to utilize 
various substrates in an in vitro synthesis. 
Using dAPPP and TPPP in the substrate, 
base-paired homopolymers of deoxyadenylic 
acid and thymidylic acid (designated as poly 

u See J. S. Krakow, H. O. Kammen, and E. S. 

Canellakis (1961). 

,; See P. Berg, H. Fancher, and M. Chamberlin 

(1963). 

7 See S. Lee-Huang and L. F. Cavalieri (1963), 

and L. F. Cavalieri (1963). 



[dA -(- T]) are synthesized without an ap- 
preciable lag period. Since each DNA 
strand of the product is homopolymeric 
(proved by nearest-neighbor analysis), it is 
clear that poly (A -f- U) is the template- 
primer. No evidence is found for the for- 
mation of a complex of poly A with poly T 
or of poly U with poly dA. Moreover, 
neither single-stranded poly A, nor single- 
stranded poly U, nor triple-stranded poly- 
ribotides can act as primer-template for 
DNA synthesis; using poly (A -f- U) as 
primer-template, no synthesis occurs with 
either JAPPP or TPPP alone. Finally, even 
after a one-fold synthesis, most — if not all — 
of the original poly (A -f- U) is present in 
original form. Consequently, no DNA-RNA 
double-helix hybrid molecules seem to be 
formed, no complete strand separation of 
poly (A + U) appears to be involved in 
synthesizing poly (dA + T), and both DNA 
strands need to be synthesized simultane- 
ously. Other helical, double-stranded, 
homopolymeric polyribotides can also be 
used with DNA polymerase to produce 
other double-stranded. homopolymeric 
DNAs (one is probably dGdC). 

Such in vitro studies may have important 
bearings upon the fate of RNA in vivo. Is 
RNA (genetic or nongenetic) ever used in 
vivo to make DNA (genetic or nongenetic)? 
To what extent does nucleic acid replication 
occur in vivo without complete strand sep- 
aration and the formation of template-prod- 
uct hybrid molecules? 



SUMMARY AND CONCLUSIONS 

DNA can be synthesized in vitro. Extended synthesis can be obtained with pre-existing 
single-stranded DNA; the 5'-triphosphates of deoxyadenosine, deoxycytidine. deoxy- 
guanosine, and thymidine; Mg++ ions; and a DNA polymerase. In making the prod- 
uct, the polymerase takes directions from pre-existing single-stranded DNA acting as a 
template. //; vitro, strands lengthen at their nucleoside end, the strand lengthened serv- 
ing as a primer. No lag period occurs in syntheses requiring base-pairing; the amounts 
of new DNA produced by these syntheses are either limited (when a pre-existing strand 



2!M) CHAPTER 21 

is lengthened slightl) ) or extensive (when essentially wholly-new strands arc produced). 
Alter .1 lag period. /.. coli I)N A polymerase can synthesize dA I and dGdC <lc novo; 
that is. m the absence of pre-existing DNA; call thymus DNA polymerase cannot 
catalyze de novo DNA synthesis. 

Nearest-nucleotide-neighbor analysis reveals that complementary DNA strands are 
synthesized in vitro in opposite directions, and that each type of natural DNA has 
unique dinucleotide sequences not predictable from its base composition. Nucleotide 
sequences, hence, do not occur at random in vivo, as strikingly exemplified in a "natural 
dAT" DNA found in a crab. 

I he DNA product of an extensive synthesis in vitro closely resembles natural DNA 
in primary and secondary structure and in other physical and chemical respects. Conse- 
quently, this in vitro DNA synthesizing process is considered to be a biological process. 
These results also support the Watson-Crick structure of chromosomal DNA and its 
replication, after strand separation, through the formation of complementary strands. 

//; vitro, E. coli DNA polymerase is reported to use double-stranded RNA to syn- 
thesize complementary double-stranded DNA without involving complete strand 
separation. 

REFERENCES 

Berg. P.. Fancher. H.. and Chamberlin, M.. "The Synthesis of Mixed Polynucleotides 
Containing Ribo- and Deoxyribonucleotides by Purified Preparations of DNA 
Polvmerase from Escherichia coli," in Informational Macromolecules, Vogel, H. J., 
Bryson, V.. and Lampen, J. O. (Eds.), New York: Academic Press, 1963, pp. 
467-483. 

Bessman, M. J., "The Replication of DNA in Cell-Free Systems," Chap. 1, pp. 1-64, in 
Molecular Genetics, Part I, Taylor, J. H. (Ed.), New York: Academic Press, 
1963. 

Bollum, F. J., " 'Primer' in DNA Polymerase Reactions," Progr. Nucleic Acid Res., 
1:1-26. 1963. 

Bollum, F. J.. "Studies on the Nature of Calf Thymus DNA-Polymerase Products," 
Cold Spring Harb. Sympos. Quant. Biol., 28:21-26, 1964. 

Burton. K., Lunt. M. R., Petersen, G. B., and Siebke, J. C, "Studies of Nucleotide 
Sequences in Deoxyribonucleic Acid," Cold Spring Harb. Sympos. Quant. Biol., 
26:27-34, 1964. 

Cavalieri, L. F., "Nucleic Acids and Information Transfer," J. Cell. Comp. Physiol., 
62 (Suppl. 1 to No. 2) :1 11-122, 1963. 

Habermann. U.. Habermannova. S.. and Cerhova, M., "The Distribution of Nucleotides 
into Pvrimidine and Purine Nucleotide Clusters in the Polynucleotide Chain of 
DNA from Escherichia coli C," Biochim. Biophys. Acta. 76:310-311, 1963. 

Romberg. A.. Enzymatic Synthesis of DNA, New York: J. Wiley & Sons, 1962. 

Romberg. A.. Bertsch, L. L., Jackson. J. F.. and Rhorana. H. G., "Enzymatic Synthesis 

of Deoxyribonucleic Acid, XVI. Oligonucleotides as Templates and the Mechanism 
o! Their Replication," Proc. Nat. Acad. Sci., U.S., 51:315-323. 1964. 

Krakow. J. S.. Kammen, H. O.. and Canellakis, E. S., "The Incorporation of Ribo- 
nucleotides into Terminal Positions of Deoxyribonucleic Acid." Biochim. Biophys. 
Acta. 53:52-64. 1961. 

Lee-Huang, S.. and Cavalieri, L. F., "Polyribonucleotides as Templates for Polydeoxy- 
ribonucleotides," Proc. Nat. Acad. Sci.. U.S.. 50:1116-1122, 1963. 



Replication of DNA in Vitro 291 

Sinsheimcr, R. L., •'Single-Stranded DNA," Scient. Amer.. 207 (No. 1): 109-1 16, 1962. 

Spencer, J. H., and Chargaff, E., "Studies on the Nucleotide Arrangement in Deoxyribo- 
nucleic Acids. VI. Pyrimidine Nucleotide Clusters: Frequency and Distribution 
in Several Species of the AT-Type," Biochim. Biophys. Acta, 68:18-27, 1963. 

Richardson, C. C. Schildkraut, C. L.. and Romberg, A.. 'Studies on the Replication 
of DNA by DNA Polymerases," Cold Spring Harb, Sympos. Quant. Biol "»8 9 19 

1 964. 

See Supplement V. A list of references is given at the end of Dr. Romberg's Nobel 
Prize lecture. 

QUESTIONS FOR DISCUSSION 

21.1. Has this chapter dealt with genetics? Explain. 

21.2. Which single experiment described in this chapter would you consider the most 
important? Why? 

21.3. Differentiate between the action of splenic phosphodiesterase and snake venom 
diesterase. 

21.4. What are the requirements for the in vitro synthesis of DNA to proceed as a 
limited reaction? To proceed extensively? 

21.5. What effect does the absence or presence of pre-existing DNA have upon DNA 
strand formation in vitro? 

21.6. List the evidence that synthesis of DNA in vitro represents a biological process. 

21.7. Does DNA polymerase from E. coli take directions only from E. coli DNA? 
Explain. 

21.8. Does strand separation occur during an extended synthesis of DNA in vitro? 
Explain. 

21.9. Of what significance is the nearest-nucleotide-neighbor analysis? 

21.10. What is wrong with Figure 8 in Supplement V? 

21.11. If substrate depletion is prevented, what would you expect to happen to the 

A + T 

ratio when an extended synthesis is permitted to proceed for several 
C + o 

hours? 

21.12. How can you explain that an extended synthesis using dGdC as primer-template 
usually yields a product richer in G than C? 

21.13. List the similarities and differences between the DNA polymerases isolated from 
E. coli and calf thymus. 

21.14. What is meant by the statement that newly synthesized DNA is covalently linked 
to the primer but is not covalently linked to the template? 

21.15. State three different procedures which you might use to synthesize dGdC in vitro. 



( 'hapter 22 

CLONES; TRANSFORMATION; 
STRAND RECOMBINATION IN VITRO 







i k present understanding of 
the mechanisms involved in 
the biological replication of 
DNA in vivo (Chapter 21 ) has been very 
significantly advanced by experiments using 
the DNA as well as the DNA polymerase of 
bacteria. Electron microscopic examination 
of bacteria reveals a nuclear region within 
which it is possible to detect a chromosome- 
like structure composed of DNA uncom- 
bined with basic protein (p. 275). The 
DNA within the bacterial nuclear body is 
similar to typical chromosomal DNA in the 
following ways: A = T and C - G; primary 
and secondary organization; mechanism of 
synthesis; and molecular integrity. There- 
fore, it seems justified to consider bacterial 
DNA as being primarily chromosomal DNA, 
and, despite the chemical simplicity of this 
structure, to use the term "chromosome" in 
discussing bacteria. 

Since bacteria contain chromosomal DNA, 
one expects them also to contain chromo- 
some-type genes according to the hypothesis 
— for which much indirect support has al- 
ready been presented — that DNA is genetic 
material. How suitable are bacteria as ex- 
perimental material for the study of genetics? 
The electron microscope reveals that each 
Escherichia coli cell contains one to four 
nuclear areas — usually two or four (Figure 
22-1 ) — and that no nuclear membrane is 
present. Although the morphological mech- 
anism of nuclear division is still unknown, 
a duplication of DNA occurs for each nu- 
292 



clear body division, and it can be concluded 
that daughter nuclear bodies are genetically 
identical, just as they are after a typical 
mitosis. 

Clones 

After nuclear-body replication, the bacte- 
rium divides to produce daughter bacteria. 
This method of increasing bacterial cell num- 
ber is an asexual process called vegetative 
reproduction. Starting with a single bac- 
terium, continuous vegetative reproduction 
results in a population of cells called a clone; 
barring mutations, all members of a clone 
are genetically identical. If mutation occurs 
during clonal growth, the mutant is trans- 
mitted to all the progeny of the mutant cell, 
thus producing a genetically mosaic clone 
whose proportion of mutant individuals 
varies, depending upon the time the muta- 
tion occurred and the relative reproductive 
potential of mutant and nonmutant cells. 
(All cells of a sexually-reproducing organ- 
ism are also colonal in origin, except for 
fertilization and meiosis and its products, 
so that multicellular organisms can also be 
mosaic for a mutant.) 

Consider the characteristics of bacteria 
and their clones significant for a study of 
mutation. The ease and speed with which 
large populations of bacteria can be ob- 
tained are of great advantage in mutation 
studies. For example, under appropriate 
culture conditions, E. coli divides about once 
each half hour; in fifteen hours after thirty 
successive generations have taken place one 
cell produces a clone containing about ten 
billion (K) 1 ") individuals. The number of 
E. coli produced from a single cell after n 
generations (or t hours) can be calculated 
by the expression 2" (or 2 2t ) (Figure 22-2). 
Space is no problem in working with bac- 
teria since 10 10 individuals can readily be 
grown in liquid broth in an ordinary test 
tube. 

However, the small size of bacteria is a 



Clones; Transformation; Strand Recombination in Vitro 



293 





B 



figure 22-1 (above). Electron microscope 
photographs of Escherichia coli. A. Whole 
cells in which nuclear bodies are revealed as 
less dense areas. Original magnification 3000X, 
present magnification about 12,000X- (Cour- 
tesy of E. Kellenberger.) B. Thin section 
showing nuclear bodies and the fine DNA- 
containing fibers within them. Original mag- 
nification 10,000X. Present magnification 
about 15,000X. (Courtesy of W. H. G. 
Schreil.) 



figure 22-2 (right). The geometric increase 
in the number of bacteria due to vegetative 
reproduction. 




V-l HOUR 





A A A A 

00000000 "■ 



1 HOUR 



HOURS 



n 3t 
2 = 2 



t HOURS 



294 



< n\i' 1 1 i< 22 




i id Ri 22-3. Discrete clones, grow- 
ing on agar nutrient in a petri dish, 
obtained by plating a dilute culture 
of bacteria. 



i IGURE 22-4. Separate bacterial 
clones obtained by the streaking 
method. {Courtesy of TV. E. Mele- 
chen. ) 




handicap in detecting phenotypic changes 
due to mutation. Mutants that change the 
morphology of bacteria must be detected by 
microscopic examination. Unfortunately, 
individual bacteria show relatively few clear- 
cut morphological variations — traits such as 
si/e. shape, capsule, pigment, and the pres- 
ence or absence of flagella. The only mu- 
tants detected then are those which alTcct 
the relatively few morphological traits to a 
measurable degree, seriously limiting the de- 
tection of mutants by examination of indi- 
vidual bacteria. 

Nevertheless, if a suspected mutant is 
found under the microscope, it is essential 
to isolate it and. from the members of the 
clone it produces, determine whether or not 
the new trait appears in the offspring. One 
of the several methods of isolating an indi- 
vidual bacterium is the tedious though exact 
procedure of removing a single individual 
from a bacterial culture with a micromanip- 
ulator and placing it in a fresh culture 
medium. Single bacteria can also be ob- 
tained by two indirect methods, less exact 
but quicker. If the bacteria are growing 
in a liquid medium, the culture can be suffi- 
ciently diluted so that a sample of it contains 
relatively few individuals. When this sample 
is poured onto the surface of a petri dish 
containing a nutrient agar medium, the in- 
dividual cells are distributed on the medium 
at random and usually so spaced that the 
visible clone which each cell later produces 
is discrete (Figure 22-3). As an alterna- 
tive, a small amount of a broth culture can 
be picked up in a sterile inoculating loop and 
the bacteria spread by streaking the loop 
across the surface of fresh nutrient agar 
medium (Figure 22-4). At some places 
on the medium, single bacteria will be de- 
posited sufficiently far apart to give rise to 
separate colonies. Which of these methods 
is used depends upon the precision required. 

The study of the individual bacterium is 
presently restricted to morphological varia- 



Clones; Transformation: Strand Recombination in Vitro 



295 



tion, since it is not yet feasible to make 
physiological and biochemical studies on 
such a microscopic scale. One can. how- 
ever, make use of the fact that, barring 
mutation, clones are composed of genetically- 
identical individuals. Genetically-different 
clones can show phenotypic differences in 
the size, shape, and color they produce on 
agar. Genetically-different clones can also 
respond differently to various dyes, drugs, 
and viruses. Therefore, one can also estab- 
lish the genotype of a single bacterium from 
the phenotype of the clone it produces. 
E. coli is easily cultured since it can grow 
and reproduce on a simple, chemically-de- 
fined food medium. Strains which grow on 
such a basic, minimal medium are considered 
to be prototrophic, or wild-type, capable of 
synthesizing the numerous metabolic com- 
ponents of the cell not supplied in the me- 
dium. In this respect prototrophs of E. coli 
or other bacteria are similar to the wild-type 
of Neurospora which also grows on a min- 
imal, chemically-defined medium. It is not 
surprising, then, that in bacteria (and also 
in Neurospora) the richest source of mu- 
tants comes from the study of the biochem- 
ical variations which occur in different 
clones, particularly those involving changes 
in nutritional requirements. For numerous 
mutants to grow and reproduce — whether 
they arise spontaneously or after treatment 
with physical or chemical mutagens — one or 
more of a variety of chemical substances 
must be added to the basic medium. For 
example, one strain of E. coli requires the 
addition of the amino acid threonine to the 
minimal medium; another strain requires the 
amino acid methionine. Nutritionally de- 
pendent strains whose growth depends on a 
supplement to their basic food medium are 
said to be auxotrophic. 

Transformation 

As characterized by clonal phenotypes. 
Pneumococcus (Diplococcus pneumoniae) 



occurs in a number of uenctic types. One 
type, S. produces a colony whose smooth 
surface is directly related to the capsule of 
polysaccharide material each bacterium pos- 
sesses. Another type of colony, R, has a 
rough surface because its bacteria lack this 
polysaccharide capsule. Moreover, several 
types of S colonies can be distinguished from 
each other because they differ antigenically; 
that is, different antisera can be obtained 
which specifically cause the clumping of each 
different type of S. R cells also occur in 
several different antigenic types, and antisera 
can also be produced which will clump them. 

In one experiment, a large number of R 
cells is placed in a nutrient broth containing 
corresponding anti-R serum. 1 When growth 
continues, clumps of agglutinated R cells 
settle to the bottom of the test tube, and the 
initially cloudy supernatant fluid becomes 
clear. If this supernate is plated on nutrient 
agar, the bacteria still present form typical 
R colonies; any mutation from R to S must 
occur rarely, for it is not detected with this 
particular technique. 

When the same experiment is performed 
with heat-killed (65 : C for 30 minutes) S 
cells also present in the nutrient broth, nu- 
merous clones of S type appear on the agar 
after plating the supernate. This S pheno- 
type is stable and clearly the result of a 
genetic change. Therefore, one is led to 
assume that the heat-killed S cells are acting 
as a mutagen in the genetic transformation 
of R to S cells. What is most surprising is 
that the type of S mutant produced is always 
identical to that of the heat-killed bacteria 
presumably acting as mutagen. In this ap- 
parently unique situation, the mutagen acts 
specifically to produce mutations in only one 
predictable direction (to one S type) rather 



1 The following account is based upon experiments 
of F. Griffith, of M. H. Dawson and R. H. P. Sia. 
and of O. T. Avery, C. M. MacLeod, and M. 
McCartv (1944). 



:m 



CHAPTER 22 



than in several directions (to two or more 

S types i. 

To determine the chemical nature of the 
mutagen involved, the transforming capacity 
oi different tractions of the heat-killed S 
bacteria is tested. Fractions containing 
either the polysaccharide coat, protein, or 
RNA are completely inactive; only the frac- 
tion containing DNA has the ability to trans- 
form. The purest DNA extracts retain the 
full transforming ability, even though they 
contain less than .02% protein or are treated 
with protein-denaturing agents or proteolytic 
enzymes. Chemical analyses and other tests 
(serological, electrophorctic, ultracentrifugal, 
and spectroscopic) also indicate that the ac- 
tive DNA is not detectably contaminated by 
either protein, unbound lipid, or polysac- 
charide. RNase has no effect on the trans- 
forming capacity of a purified DNA fraction, 
but the transforming factor is completely de- 
stroyed by DNase, showing that transforma- 
tion requires DNA in highly polymerized 
form. 

As revealed from its X-ray diffraction 
pattern, transforming DNA has the double- 
stranded configuration of chromosomal 
DNA. Since pure DNA can be used to 
transform, no contact need be made between 
the cell acting as DNA donor and the one 
acting as recipient. Moreover, genetic trans- 
formation does not involve the mediation of 
a virus. Therefore, beyond any reasonable 
scientific doubt, DNA alone must be the 
transforming agent. 

Transformation can occur in either direc- 
tion (A^^A'), and, in bacteria, any chro- 
mosomal gene can be transformed. Type A 
cells can be transformed to an A' type which, 
in turn, provides increased amounts of A'- 
DNA capable of transforming other A cells 
to A'. So, the DNA extracted from trans- 
formed bacteria provides increased amounts 
of the same transforming principle. 

One transforming principle (A') can 
transform bacteria having any one of several 



alternative phenotypes (for example, A or 
A"). If the A'-DNA, obtained from A-type 
bacteria transformed to A', is then used to 
transform bacteria of a third genotype. A", 
only A' transformants are found — the only 
transformations produced are those involv- 
ing the genes of the immediate donor. This 
result demonstrates that transformation pro- 
duces a transmissible alteration based upon 
the loss of host genetic material apparently 
at the same time as the new genetic material 
is acquired — not the simple addition of par- 
ticular genetic material to the genotype. 
Thus, the genetic change in transformation 
is of a replacement type. 

The fate of transforming DNA can be 
traced - by labeling its phosphate groups 
with radioactive P 32 . At various times after 
exposure to such labeled DNA, one portion 
of the treated bacteria is killed and analyzed 
for the presence of P :i - in its DNA, whereas 
another portion is tested to determine trans- 
formation frequency. Only after bacteria 
have been exposed to the DNA extract for 
a suitable period of time is the labeled DNA 
found in the extract containing the host's 
chromosomal DNA. Moreover, the fre- 
quency with which the host cell is trans- 
formed is directly proportional to the amount 
of labeled DNA so incorporated. 

The preceding results demonstrate that the 
transforming DNA actually enters the bac- 
terium and replaces a segment of the host's 
chromosomal DNA { (Pneumococcus con- 
tains about 6 times 10 fi deoxyribotide pairs 
per nuclear body), after which the newly- 
introduced material replicates as a normal 
part of the chromosome. Since it is DNA 
which alone carries the genetic information 
for transformation, genetic transformation 
provides direct and a inclusive evidence that 
DNA is genetic material. Accordingly, 

- Based upon work of L. S. Lerman and L. J. 
Tolmach (1957). 

3 See H. Ephrussi-Taylor (1951) for specific evi- 
dence on the latter observation. 



Clones; Transformation; Strand Recombination in Vitro 



297 



chromosomal DNA contains the chemical 
units of the genetic material; all other trans- 
formation studies support this conclusion. 

We should now re-examine the assump- 
tion, made earlier in this chapter, that trans- 
formation involves mutation. Because the 
first transformation studies seemed to in- 
volve novel, rare changes in the genetic ma- 
terial, these were called mutations (p. 149). 
We now know that transformation involves 
replacement of one segment of genetic ma- 
terial by another, that only a shuffling of 
already-existent genes occurs and not a new 
type of genetic material. Moreover, genetic 
transformation has been found not only in 
Pneumococcus but in Hemophilus, Xantho- 
monas, Salmonella, Bacillus, Neisseria, Esch- 
erichia, and other organisms as well. In 
Neisseria, DNA is regularly liberated (into 
the slime layer) by the cells which undergo 
self-digestion, or autolysis, in aging cultures; 
such DNA is effective in transformation, as 
is the DNA obtained from penicillin-sensi- 
tive pneumococci disintegrated or lysed 
after treatment with penicillin. Using dif- 
ferent genetically-marked pneumococci, it is 
found 4 that genetic transformation — due to 
DNA liberated from one strain transforming 
members of the other strain — occurs spon- 
taneously in the living mouse host. 

Two lines of the human cell strain D98S 
maintained in vitro are genetically differen- 
tiated by the presence or absence of the 
enzyme, inosinic acid pyrophosphorylase 
(IMPPase). Since the spontaneous muta- 
tion frequency from IMPPase-negative to 
-positive is found to be less than one cell 
in 10 7 , and since a culture medium can be 
employed which completely prevents cells 
of the negative line — but not those of the 
positive line — from forming colonies, it is 
possible to detect as few as one genetic trans- 
formant per 10 7 cells. Treatment of the 
IMPPase-negative line with DNA isolated 

4 By E. Ottolenghi and C. M. MacLeod (1963). 



from IMPPase-positive cells results in the 
appearance of IMPPase-positive, genetically- 
transformed cells at rates as high as 4 times 
10 4 transformations per recipient cell (one 
transformation per 2500 treated cells). 
Such rates offer clear proof that genetic 
transformation of human cell lines occurs 
under experimental conditions." 

Not only is transformation widespread, 
but a given type can occur with a frequency 
as high as 25%. Such results demonstrate, 
of course, that transformation is not rare. 
Because genetic transformation is not rare 
and does not produce novel genotypes, it 
should not be considered a type of mutation. 
Accordingly, just as with segregation, inde- 
pendent segregation, crossing over, and fer- 
tilization, it is probably best to consider ge- 
netic transformation as another mechanism 
for genetic recombination. 

As determined from bacterial studies, the 
complete transformation process requires a 
series of discrete stages, as follows: 

Cell competence. During certain periods 
in cell division or in the growth of a bacterial 
culture, transformation does not occur; in 
other periods the cells are competent to 
react. 

Binding the transforming DNA. When 
bacteria are in a competent stage, the trans- 
forming DNA, transiently bound to the cell 
at first, can be removed by several methods 
including exposure to DNase, before the 
DNA is permanently bound. 

Penetration of transforming DNA. Per- 
manently-bound DNA is considered to have 
penetrated the recipient bacterium. It 
should be noted that the success of trans- 
formation is inversely related to the thick- 
ness of a polysaccharide coat which probably 
acts as some kind of barrier to binding or 
penetration. When transforming DNA is 
fragmented sonically, the newly formed DNA 

5 According to W. Szybalski. E. H. Szybalska, and 
G. Ragni (1962); see E. H. Szybalska and W. 
Szybalski (1962). 



298 



( II M'TI.R 22 



particles have a molecular weight of less 
than 4 times H>\ which is not sufficient 

to penetrate. Only high-molecular-weight 
DNA penetrates. 

These tacts should be considered with re- 
gard to the circumstances under which DNA 
uptake occurs in mammalian tissue culture. 
In this case, DNA enters by phagocytosis 
which occurs only when the DNA adheres 
to a suitably large non-DNA particle. Pino- 
cytosis, similar to phagocytosis, is another 
process by which materials can enter ordi- 
nary cells. Although pure nucleic acids are 
not pinocytosed, protein is. However, if 
pure nucleic acid is mixed with protein, 
pinocytosis is stimulated, and the nucleic 
acid is carried into the cell with the protein. 
Perhaps the penetration of high-molecular- 
weight DNA into bacteria is dependent upon 
the presence of sufficient contaminating ma- 
terial capable of stimulating pinocytosis or 
some other mechanism for DNA penetra- 
tion. Whatever the precise method by which 
transforming DNA penetrates, it is found 
that relatively short sequences of DNA will 
enter microbial cells if sufficient protein is 
also present. 

The bacterial surface contains a finite 
number of sites which act as receptors for 
DNA. Since non-transforming DNA (such 
as DNA from a widely separated genus) can 
also penetrate readily, receptor sites can be 
saturated by nontransforming DNA, thereby 
preventing the penetration of transforming 
DNA. 

Synapsis. Alternatives of the same trait 
— for example, resistance and sensitivity to 
streptomycin, or auxotrophy and prototrophy 
for a particular nutrient — can be found in 
different species of bacteria. Since it is a 
reasonable assumption that the same type 
i)| gene (and its alternatives) performs the 
same or similar functions in different species, 
interspecific transformations ought to be 
possible. Although this result has been 
achieved, in any given case the interspecific 



transformation is usually less frequent than 
the intraspecific one. Moreover, the trans- 
formation frequency is actually lower and 
not due to a delay in phenotypic expression 
which occurs in interspecific (but not in 
intraspecific) transformation. That inter- 
specific transformation does take place favors 
the idea that the transformed locus is nor- 
mally part of the genotype of both species. 
The relative infrequency of interspecific 
transformations is, therefore, not due to in- 
competence of the recipient cell or a failure 
of the foreign DNA to bind to or penetrate 
the recipient. 

The transforming capacity of already- 
penetrated DNA may depend not only upon 
the homology of the loci transformed but 
upon the nature of the genes adjacent to those 
undergoing transformation. These neigh- 
boring genes might influence transformation 
by their effect upon the synapsis of the trans- 
forming DNA with the corresponding region 
of the host's genetic material. In intraspe- 
cific transformation, the loci adjacent to 
those transformed are very probably homol- 
ogous in transformer and host, so that syn- 
apsis between the two segments can occur 
properly; in interspecific transformation, 
these loci are likely to be nonhomologous 
and, therefore, may often fail to synapse or 
act to prevent synapsis. 

Integration. Even if the hypothesized 
synapsis occurs properly between host and 
transforming DNA, some process has yet 
to occur by which the host gene — whose 
transformation is being followed — is lost 
from the chromosome, and the donor's locus 
becomes an integral part of it. Some under- 
standing of the mechanism of this final stage 
in transformation may be gained from a 
study of transformation frequency. First of 
all. different loci transform intraspecifically 
with different frequencies. Using genes that 
transform with suitably high frequencies, we 
are able to study the frequency of double 
transformations, that is, the frequency with 



Clones; Transformation; Strand Recombination in Vitro 



299 



which bacteria are transformed with respect 
to two markers present in the donor DNA. 
In several cases (for example, penicillin- and 
streptomycin-resistance), the frequency of 
doubly-transformed bacteria is approxi- 
mately equal to (actually somewhat less 
than) the product of the frequencies for the 
single transformations. Such results prob- 
ably mean that the transforming DNA car- 
ries the two loci either on separate particles 
or in widely separated positions on the same 
particle. On the other hand, the markers 
for streptomycin-resistance and mannitol- 
fermentation are transformed together with 
a frequency (.1%) which is about 17 times 
that expected from the product of the fre- 
quencies of the single transformations 
(.006%). This result implies that these 
two genetic markers are located on the same 
transforming particle; that is, they seem to 
be reasonably close together in the same 
bacterial chromosome. 

If two loci are closely linked, how can we 
explain the occurrence of single and double 
transformations for them? Because of frag- 
mentation during extraction, a given pene- 
trating DNA particle may not always have 
the same composition relative to the two 
markers; it may sometimes carry only one 
and, at other times, may carry both of these 
markers. The effect of reducing the particle 
size of penetrating DNA upon the fre- 
quencies of single and double transforma- 
tions can be tested. When particle size is 
reduced by DNase or sonic treatment, one 
expects — according to the present hypothe- 
sis — the particles sometimes to be broken 
between the two markers, reducing the rela- 
tive frequency of the double transformation 
and increasing the relative frequencies of the 
single transformations. When the particle 
size is reduced, the overall rate of trans- 
formation is lower, as expected. No change 
is found, however, in the ratio of double to 
single transformations, implying that the two 
markers are so closely linked, they are rarely 



separated when particles are fragmented. 
Accordingly, it seems that the penetrating 
particles must usually carry both markers, 
or neither, and the failure to obtain 100% 
double-transformations from the former type 
must be because only a small portion of a 
penetrating, synapsing particle is integrated. 

Integration of a portion of a synapsed 
particle can occur in two possible ways (Fig- 
ure 22-5): One involves copy-choice (Fig- 
ure 22-5B) in which a daughter chromo- 
some is formed by the alternate use of the 
host chromosome and the donor DNA as a 
template. When completed, the daughter 
chromosome is exactly like the original chro- 
mosome except for the daughter segment 
formed with transforming DNA as the tem- 
plate. One expects the recombinant chro- 
mosome produced by the copy-choice 
method to contain all newly-synthesized 
DNA. 

The second method involves breakage and 
exchange of the kind that takes place in 
chromosomal rearrangement or in crossing 
over. In this case (Figure 22-5A), 
"breaks" have to occur on each side of the 
marker being integrated, so that a "double 
crossover" (p. 134) is produced. Although 
double crossovers within a short distance 
are expected to be extremely rare between 
two homologous chromosomes of higher 
organisms, this kind of exchange can occur 
under special circumstances and may be pos- 
sible between the chemically less complex 
chromosome of bacteria and the shorter, 
synapsed segment of transforming DNA. 
Linkage of transforming DNA to host mark- 
ers does not require DNA synthesis in the 
region involved, ,! although the integrated 
segment — which must be at least 900 nu- 
cleotide pairs long — appears to replicate in 
synchrony with the host DNA. Experiments 
with labeled DNA show that in transforma- 
tion single-stranded donor DNA is inserted 

«See M. S. Fox (1962). 



300 







Cl 




b 


c 




c 


d' 




d 


e' 




e 


P 




f 




g 






h 



n 11 




CHROMOSOME 



TRANSFORMING 
SEGMENT 















CHAPTER 22 




a 




a 




b 




b 


c' 


c 


c' 




c 


d' 


d 


d' 




d' 


e' 


e 


e' 




e' 


P 


f 


P 




f 




g 




g 




h 




h 




* 




INTEGRATION COMPLETED 




+ 




a 




a 


a 




b 




b 


b 


c 




c 


c' 




c 


c 


d' 




d 


d' 




d' 


d 


e' 




e 


e 




e' 


e 


P 


! : 


f 


P 




f 


f 




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g 




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■ 


h 


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REPLIC 


DAUGHTER 








CI 


HROM 


OSOME 





figure 22—5. Postulated mechanisms for the incorporation of a segment of genetic 
information into a host chromosome. A. Breakage method; B. Copy-choice method. 



into the DNA of the host, and that probably 
either strand of the donor DNA could be 
used this way. 7 Thus, the evidence favors 
the breakage hypothesis of integration. 
It is important to note that the portion 



■ See O. H. Siddiqi (1963), and M. S. Fox and 
M. K. Allen (1964). 



of penetrating donor DNA which is not in- 
tegrated, obviously, is also not retained or 
conserved as chromosomal genetic material. 
If integration occurs by copy-choice, the 
transforming segment is not conserved 
either; if integration occurs by "breakage- 
exchange." the replaced DNA segment is 
not conserved. 



Clones; Transformation; Strand Recombination in Vitro 



301 



Strand Recombination in Vitro 

Heating chromosomal DNA denatures it by 
causing strand separation. After quick cool- 
ing, the resultant single strands of denatured 
DNA can fold forming a considerable num- 
ber of complementary base pairs between 
bases at different levels of the single strand. 
It should be noted that, under certain con- 
ditions, homopolymers of RNA containing 
A, U, C, or inosinic acid are capable of base- 
pairing after folding, thus forming regular 
double-helical structures. To understand 
de novo synthesis and limited reaction in 
vitro it may be important to learn the ex- 
tent to which intrastrand base-pairing oc- 
curs in DNA homopolymers containing C 
or G. After pneumococcal DNA is heated 
for ten minutes at 100° C, all strands are 
essentially single and all H bonds, broken; 8 
this denaturation, called melting, occurs 
sharply at 71° C for dAT and at 83° C for 
dGdC. When DNA denatured by heat is 
cooled slowly, only about 70% renaturation 
occurs for native DNA, although, as ex- 
pected, dAT and dGdC apparently show 
100% renaturation. 

Renatured and native DNA differ from 
denatured DNA in several properties: 

1. Under the electron microscope, rena- 
tured DNA looks very much like na- 
tive DNA, whereas denatured DNA is 
irregularly coiled with clustered re- 
gions 

2. Renatured and native DNA have sim- 
ilar and lighter densities than dena- 
tured DNA 

3. Renatured and native DNA have about 
twice the molecular weight of dena- 
tured DNA 

4. Although all DNA has the same ab- 
sorption spectrum, renatured and na- 

8 The following account is based primarily on work 
reported by P. Doty, J. Marmur, J. Eigner, and 
G. Schildkraut (1960), and J. Marmur and D. 
Lane (1960). 



tive DNA absorb less ultraviolet than 
denatured DNA. 

Several factors affect renaturation: 

1 . The concentration of DNA in a slowly 
cooling mixture. When the concentration 
of single strands is high, so is the amount of 
renaturation; when the concentration is low, 
slow cooling does not produce any substan- 
tial recombination of strands. 

2. Salt concentration. The negatively- 
charged phosphate groups of single strands 
tend to prevent union with other strands. 
This inhibition can be overcome by adding 
KC1 to the solution to act as a shield against 
the repulsion between phosphates. Conse- 
quently, within a certain range, the more 
KC1 present, the greater the amount of re- 
naturation obtained by slow cooling of 
heated DNA. 

3. The source of DNA. Assuming the 
molecular weight of native DNA to be ap- 
proximately the same in all organisms, a 
mammalian cell, which has about a thousand 
times as much nuclear DNA as a bacterial 
cell, also has about a thousand times as many 
DNA molecules. Assuming that all the 
DNA molecules within a genome differ in 
base sequence, then, in a given concentra- 
tion of denatured DNA, on the order of one 
thousand times fewer complementary strands 
are present in a sample from calf thymus 
than in one from Pneumococcus. When 
equal concentrations of denatured DNA are 
heated to 80° C, double strands are formed 
by a large fraction of the bacterial DNA, 
but by no detectable fraction of the calf 
thymus DNA. The concentration of com- 
plementary strands, therefore, is important 
in renaturation. 

DNA can be denatured in vitro by a large 
number of organic chemical substances in- 
cluding urea, aromatic compounds, and a 
variety of alcohols. This finding, however, 
does not necessarily mean that such com- 
pounds have this function in vivo, or that 



302 



( II M'lI.K 22 



thej reveal what is responsible for holding 

the strands in a DNA double helix together 
under in vivo or the usual in vitro conditions. 

Vnother physical-chemical change can 
occur when DNA is heated in vitro. As 
noted, native pneumococcal DNA has a 
molecular weight of about six million. 
When certain preparations of this native 
DNA are heated, the single strands obtained 
have a molecular weight o\' less than half 
this value. This reduction in molecular 
weight can be explained by the presence o\ 
DNase as a contaminant. Even though 
single strands in a double helix are enzy- 
matically severed by DNase, the whole com- 
plex can still retain the double-helix con- 
figuration. Once these complementary 
strands are separated by heat denaturation, 
however, the fragments of each single strand 
separate. 

As already mentioned, DNA from differ- 
ent sources and DNA particles of different 
sizes behave differently in various parts of 
the sequence leading to transformation. 
When DNA in vitro is exposed to dilute con- 
centrations of DNase, the results 9 indicate 
that single strands of the double helix are 
attacked first, and only later — when both 
strands have been attacked at reasonably 
nearby positions — is the molecule severed. 
This scission produces smaller DNA mole- 
cules which, it should be recalled, penetrate 
a host cell poorly. Even if only one strand 
of the double helix has been attacked, how- 
ever, transformation capacity declines. This 
effect is attributed partly to the failure of 
penetrant molecules to transform because 
the transforming locus or because a locus 
neeessary for synapsis or integration has 
been inactivated. 

In Pneumococcus, denatured DNA has a 
small amount of transforming ability. The 
molecular basis for this is still undetermined. 
On the other hand, the transforming ability 

1 Of L. S. Lerrrum and I.. J. Tolmach. 



of renaturcd DNA can be as much as 
509? o[ that shown by an equivalent con- 
centration o\' native DNA. An increased 
concentration of DNA plus a high ionic 
strength increase both renaturation and 
transforming ability. 

Hybrid molecules can be made by rena- 
turing a mixture of N-14 and N-15 DNA 
from E. coli. ( Recall that these synthetic 
molecules can be identified by the inter- 
mediate position they assume in the ultra- 
centrifuge tube.) Hybrid molecules can also 
be formed between single DNA strands from 
different speeies, but only if the species are 
closely related genetically (as would be sug- 
gested if they showed interspecific transfor- 
mation) and, therefore, have similar base 
sequences. Molecular hybrids are useful 
for comparing base sequences in closely- 
related organisms even when genetic recom- 
bination between them cannot take place. 

Several additional observations should be 
made: 

1 . Strand separation is accomplished by 
heat in a matter of a few minutes or less. 
One wonders if this kind of extensive strand 
separation occurs in vivo. It has been sug- 
gested that chain separation normally is pro- 
duced enzymatically through the activity of 
ravelase, or better, unravelase. 

2. The now-routine ability to separate and 
combine single strands should lead to a better 
understanding of transformation, in particu- 
lar, the mechanism of integration. 

3. The smallest recombinational unit of 
the genetic material in bacteria can he iden- 
tified as the smallest unit of DNA capable 
of being integrated or replaced in a host 
genotype in a genetic transformation. 

Although the physical and chemical prop- 
erties of the DNA product of an extensive 
synthesis in vitro closely resemble those of 
the natural DNA used as primer-template, 
and although the synthesis is considered to 
be a biological process, it has not been dem- 
onstrated that the DNA product has biolog- 



Clones; Transformation; Strand Recombination in Vitro 



303 



ical properties, that is, functions genetically 
in vivo. When transforming DNA is used 
in an in vitro synthesis, the total transform- 
ing capacity of the incubating mixture de- 
creases with time as the synthesis continues. 
It is very likely that the trace amounts of 
DNases present in the polymerase prepara- 
tion cause the loss of overall transforming 
activity by interrupting the continuity of the 
DNA strands, so that transforming capacity 
is lost by DNase action faster than it is 
gained by means of DNA synthesis. The 
biological activity of newly-synthesized DNA 
can be detected, however, by differentially 
labeling the old and the new DNA and sep- 
arately testing each for transforming capac- 
ity. 



Consequently, 10 nonradioactive DNA con- 
taining 5-bromo uracil is used as a primer- 
template to synthesize radioactive DNA with 
no 5-bromo uracil. After synthesis, density 
gradient centrifugation of the DNA provides 
a double-stranded fraction containing essen- 
tially all newly-synthesized DNA (which is 
radioactive and less dense than the DNA 
with 5-bromo uracil). Since other explana- 
tions seem to be ruled out, and this new 
DNA is found capable of transforming a 
variety of gene loci, it apparently is proved 
that biologically-active genetic material, that 
is, functionally-active genes, can be syn- 
thesized in vitro. 

10 See R. M. Litman and W. Szybalski (1963). 



SUMMARY AND CONCLUSIONS 

Genetic recombination occurs in cells of bacteria and of other organisms by means 
of genetic transformation. In bacteria, transformation involves a sequence of events in 
which competent cells transiently and then permanently bind DNA. Once bound DNA 
has penetrated, it apparently undergoes a synapsis-like process with a corresponding 
segment of the bacterial genome. Transformation is completed when a small segment 
of donor DNA becomes integrated and replaces a similar segment of the host genome. 

Transformation provides direct and conclusive evidence that chromosomal DNA 
is genetic material. In bacteria, the smallest recombinational unit of the genetic ma- 
terial is the smallest unit of DNA integrated or replaced in transformation. 

Strand separation and recombination /'/; vitro produce denatured and renatured DNA. 
respectively. 

Strong evidence has been obtained that functionally-active genes, introduced into 
bacteria via transformation, can be synthesized in vitro. 



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Roger, M., "Fractionation of Pneumococcal DNA Following Selective Heat Denatura- 
tion: Enrichment of Transforming Activity for Aminopterium Resistance," Proc. 
Nat. Acad. Sci., U.S., 51:189-195, 1964. 

Siddiqi, O. H., "Incorporation of Parental DNA into Genetic Recombinants of E. coli," 
Proc. Nat. Acad. Sci., U.S., 49:589-592, and 50:581, 1963. 

Szybalska, E. H.. and Szybalski, W., "Genetics of Human Cell Lines, IV. DNA- 
Mediated Heritable Transformation of a Biochemical Trait," Proc. Nat. Acad. 
Sci., U.S., 48:2026-2034, 1962. 



Clones; Transformation; Strand Recombination in Vitro 305 

QUESTIONS FOR DISCUSSION 

22.1. Which single characteristic of bacteria provides the greatest advantage for ge- 
netic studies? Why? 

22.2. Assuming that all members of a clone are genetically identical, could sexual 
processes have influenced the results of any of the experiments described in 
this chapter? Explain. 

22.3. Distinguish between auxotrophic and prototrophic bacteria. 

22.4. Design an experiment to test whether the dye, acriflavin, is mutagenic in E. coli. 

22.5. Compare the "chromosome" of bacteria with that of man. 

22.6. Is the use of the phrase "bacterial chromosomal DNA" justified even though 
bacteria do not contain typical chromosomes? Explain. 

22.7. What is meant by integration in genetics? Without using diagrams, describe 
the mechanisms by which it may occur. 

22.8. List those features of crossing over which are difficult to explain on a copy- 
choice basis. 

22.9. Discuss the genetic control of gene synthesis and gene degradation. 

22.10. Criticize the statement (p. 5) that genetic transmission can occur between gen- 
erations only by means of a cellular bridge. 

22.11. On what basis is transformation classified as a type of genetic recombination 
rather than as a mutation? Do you agree with this interpretation? Why? 

22.12. Devise an experiment to detect whether chain separation occurs during exten- 
sive in vitro synthesis of DNA. 

22.13. Do the studies on transformation offer any clues as to the ploidy of Pneumococ- 
cus? Explain. 

22.14. What kinds of problems would you investigate if you had a feasible method of 
studying the fate of individual cells exposed to transforming DNA? 

22.15. What do studies of genetic transformation reveal regarding the genetic nature 
of conserved and nonconserved chromosomal DNA? 

22.16. Redraw Figure 22-5 showing hypothetical base sequences in double-stranded 
DNA. Has your drawing any bearing on your answer to question 22.8? 
Explain. 

22.17. How can you explain the finding (p. 299) that the frequency of double trans- 
formations for certain markers is sometimes somewhat less than the product of 
the frequencies of the single transformations? 

22.18. Interspecific transformation in bacteria is rare or absent when the relative G + C 
contents of host and donor differ. When the G + C contents are the same, 
donor-host hybrid DNA's can form even when interspecific transformation is 
rare. Discuss the relative values of G + C content, hybrid DNA formation, 
and interspecific transformation in taxonomic studies of bacteria. 



Chapter 23 

BACTERIAL MUTATION 
AND CONJUGATION 

Mutation 

Practically all of the mutants produced after 
treatment with a mutagenic agent have non- 
adaptive or detrimental phenotypic effects 
(Chapter 16). The detriment produced by 
these mutants clearly is not dependent upon 
the mutagen's continued presence in the en- 
vironment. Similarly many of the rare mu- 
tants that increase adaptability continue to 
be beneficial in the absence of the mutagen 
which induced them. On rare occasions, 
however, a mutagen (like X rays) produces 
a mutant with an adaptive advantage in the 
presence of the mutagen (for example, re- 
sistance to the genetic or nongenetic detri- 
mental effects of X rays). Is such an adap- 
tive mutant produced by chance, or is it a 
special genetic response elicited by the mu- 
tagen? The same question can be raised 
about adaptive mutants that occur "spon- 
taneously." Are these mutants produced as 
an adaptive genetic response to unidentified 
factors in the environment? 

This general problem can be illustrated 
with a particular strain of E. coli which ap- 
parently has never been exposed to the drug 
streptomycin. If such a strain is plated 
onto an agar medium containing this drug, 
almost all the individuals will not grow and, 
there lore, will not form colonies. These in- 
dividuals are streptomyt insensitive. How- 
ever, about one bacterium in ten million 
does grow on this medium and forms a 
colony composed o\' streptomycin-resistant 
individuals, the basis for this resistance 
306 



clearly being transmissible. Is the adaptive, 
resistant mutant produced in response to the 
streptomycin exposure, with the streptomy- 
cin acting as a directive mutagenic agent? 
Or, do streptomycin-resistant mutants occur 
in the absence of streptomycin, sponta- 
neously, with the streptomycin acting only 
as a selective agent to reveal the prior oc- 
currence (or nonoccurrence) of resistant 
mutants? Or, are both explanations true? 
Restating the problem more generally, we 
ask whether mutants adapted to a treatment 
are postadapted (having arisen after treat- 
ment), pre adapted (having already been 
present before treatment), or of both types. 

Clearly, an ambiguous decision results so 
long as it is necessary to treat the individuals 
scored with what is being tested — strepto- 
mycin, in this example — for, under these 
conditions, one cannot decide whether the 
resistant mutant had a post- or preadaptive 
origin. This difficulty can be resolved. If 
streptomycin-resistant mutants are preadap- 
tive, they should occur in the absence of the 
drug and give rise to clones all of whose 
members are resistant. It should be noted 
again that the mutation to streptomycin- 
resistance is a very rare event however it 
originates. Consequently, one must grow 
about ten million clones on streptomycin- 
free agar medium and test each clone for 
streptomycin resistance by placing a sample 
of each on a streptomycin-containing me- 
dium. After this transfer, part or all of one 
sample is expected to be resistant to the 
drug. If resistance is due to a preadapted 
mutant, one can return to the appropriate 
original clone — which has never been ex- 
posed to streptomycin — and readily obtain 
other samples which prove to be resistant. 
If, on the other hand, the mutant is post- 
adaptive, additional samples of the original 
clone will have no greater chance o\' furnish- 
ing resistants than additional samples taken 
from different clones. 

Three clone-sampling procedures are 



Bacterial Mutation and Conjugation 



307 



available for testing the preadaptive or post- 
adaptive origin of mutants. The first method 
starts with growing a single (presumably) 
streptomycin-sensitive clone in liquid me- 
dium and then plating it on an agar medium 
to produce a large number of separate col- 
onies. A sample of each colony is then 
streaked across an agar medium perpendicu- 
lar to a strip in the agar containing strepto- 
mycin. Where there is no streptomycin, 
each streak of bacteria will grow on the agar 
and, if enough clones are tested, at least one 
streak will also grow in the streptomycin 
region (Figure 23-1). If streptomycin- 
resistant mutants are postadaptive in origin, 
the growth on the streptomycin-containing 
area will be sharply discontinuous, since the 
members of the clone streaked across the 
streptomycin were originally sensitives, and 
only rarely will more than one of these 
bacteria respond to streptomycin by post- 
adaptive mutation. Moreover, other sam- 
ples from the original clone will succeed in 
growing on streptomycin only to the same 
limited degree as did the first sample. If, 
on the other hand, the mutation is preadap- 
tive, the growth across the streptomycin will 
be practically as continuous as after streak- 



BAND OF STREPTOMYCIN 



RESISTANT 
CLONE 

SENSITIVE 
CLONE 



figure 23-1. Streptomycin-sensitive and 
streptomycin-resistant E. coli as determined 
by streaking individual clones. 




ing the drug with a pure clone of resistant 
bacteria or with a mixture of bacteria rich 
in resistant individuals. The proof that the 
parental clone contained a spontaneous, pre- 
adaptive, streptomycin-resistant mutant will 
be complete if other samples of that clone 
also grow readily when streaked across this 
drug. 

Considering the rarity of mutation from 
streptomycin-sensitivity to -resistance in this 
strain (one per 10 7 cells), the labor involved 
in testing the preadaptive or postadaptive 
nature of streptomycin-resistant mutants by 
this clone-sampling technique is prohibitive. 
(Nevertheless, the method has numerous 
uses in other genetic studies of bacteria.) 

The second method which can be used to 
sample clones involves replica plating. 1 As 
before, this procedure starts by spreading 
the members of a single clone on streptomy- 
cin-free agar in a petri dish. As many as 
a thousand separate colonies can form after 
the plate is incubated and by pressing this 
master plate on the top of a sheet of velvet, 
a sample of almost every colony can be ob- 
tained simultaneously. The velvet — whose 
fibers pick up 10 to 30% of each colony — 
is then used to plant a corresponding pattern 
of growth on a series of additional replica 
plates (Figure 23-2). Preliminary control 
tests show that the velvet makes several 
excellent replicas of the master plate, and 
that both streptomycin-resistant and -sensi- 
tive clones can be replicated this way. The 
first replica is made on drug-free medium, 
whereas the second and later ones are made 
with streptomycin-containing plates on 
which, obviously, only streptomycin-resist- 
ant bacteria can grow into colonies. If the 
postadaptive view is correct, the chance that 
cells from one colony will grow on two 
replicas in the presence of streptomycin is 
the same as it is for two colonies to grow 
on the same replica. In other words, the 

1 Based upon work of J. Lederberg and E. M. 
Lederberg. 



30S 



CHAPTER 23 



positions of the resistant colonies on differ- 
ent replicas arc random. II' the mutants are 
preadaptive, however, all replicas probably 
will be resistant in the same position (al- 
though some exceptions can occur if the 
velvet tails to place a portion o\' the same 
colony on every replica plate). Of course, 
one also readily finds resistant bacteria in 
the corresponding position on the master 
plate. Although this clone-sampling tech- 
nique is advantageous for many other pur- 
poses, it is still too laborious for testing the 
preadaptation or postadaptation hypothesis. 
since replicas of about ten thousand master 
plates are required to be reasonably sure of 
rinding one clonal streptomycin-resistant 
mutant. 

This difficulty can be avoided by using a 
third method for clone-sampling that in- 
volves replica-plating contiguous colonies. 
A billion or so bacteria (from a streptomy- 
cin-sensitive clone) plated on drug-free agar 
will produce small clones so closely spaced 
that they grow together and form a bacterial 
lawn (Figure 23-3A). Nevertheless, rep- 
licas of this growth can be made on strepto- 



mycin-containing agar and will show growth 
wherever drug-resistant mutants occur (Fig- 
ure 23-3B, CD). One can then turn to 
the corresponding regions on the master 
plate to obtain samples to be tested for re- 
sistance to the drug. If such samples are 
no richer in resistant mutants than samples 
from randomly-chosen sites on the master 
plate corresponding to those which are not 
mutant on any replica, the postadaptive view 
is proved. When the experiment is actually 
performed, the master plate is found to be 
much richer in mutants at replica sites that 
are mutant than at those that are nonmutant. 
Moreover, replicas tend to have mutant 
clones at corresponding positions on all 
replica plates (Figure 23-3B-D). Accord- 
ingly, most mutants are clearly preadaptive. 
Other experiments show conclusively, in the 
case of streptomycin, that almost all, if not 
all, mutants resistant to the drug are pre- 
adaptive — that is, streptomycin does not in- 
duce a detectable number of resistant mu- 
tants. Since the same results are obtained 
with the drug chloramphenicol, one can 
extrapolate and conclude that, in general, the 




figure 23-2. Separate colonies replica-plated (right) from a master plate 
(left). (Courtesy of N. E. Melechen.) 



Bacterial Mutation and Conjugation 



309 



resistant mutants on drug plates arise spon- 
taneously, prior to exposure to the drugs 
and, there jore, are preadaptive in origin. 

Large numbers of bacteria can be easily 
tested for mutations. For example, a billion 
drug-sensitive individuals can be plated on 
agar containing the drug, and the number 
of resistant mutant clones detected by count- 
ing the colonies formed; or, similarly, the 
number of mutants to prototrophy can be 
scored by plating auxotrophs on agar which 
lacks the nutrient required for their growth 
and counting the number of colonies formed. 
To give information in terms of a rate of 
mutation, however, it is necessary to state 
the number of mutants occurring per unit 
event. In multicellular organisms, mutation 
rate is usually expressed in terms of muta- 
tions per cell, per individual, or per genera- 
tion. This definition can be applied to bac- 
teria also. Thus, the mutation rate from 
streptomycin-sensitivity to -resistance in one 
particular strain of E. coli (a different strain 
from that used previously) is one per billion 
bacteria — one of the lowest mutation rates 
so far measured in any organism. 

It is sometimes desirable to express muta- 
tion rate in terms of mutations per unit time; 
for example, in describing the increase in 
mutations obtained by aging Drosophila 
spermatids or sperm (Chapter 14). In bac- 
teria, one can considerably vary the length 
of time required to complete a generation. 
For generation times between 37 minutes 
and two hours, the shorter the generation 
time, the larger the mutation rate per hour. 
When generation time is lengthened from 
two to twelve hours, the rate of mutations 
per hour is constant — each hour of delay 
increasing the number of mutants by the 
same amount. (Thus, in the two- to twelve- 
hour range, the number of mutations in- 
creases linearly per generation. ) Even when 
the generation time is extended from twelve 
hours to infinity (the nondividing cells kept 
alive in a medium which provides a source 




figure 23-3. Replica-plating a bacterial lawn 
for the detection of mutants to streptomycin- 
resistance. (After J. Lederberg and E. M. 
Lederberg. ) 



of energy), some mutations are found to 
take place. 

It becomes apparent, therefore, that mu- 
tation rate is best defined as the chance of 
a mutation per cell {or individual) per unit 
time. When, however, each of the division 
cycles or generations requires the same 
length of time (as would be true for bacteria 
under optimal environmental conditions), 
mutation rate is usually measured with one 
generation as the unit of time. 

Conjugation 

Genetic recombination in bacteria may occur 
as a result of genetic transformation. The 
transformation process has two features 
hitherto unencountered in discussions of ge- 
netic recombination in multicellular organ- 
isms: 

1. The donor DNA enters the host bac- 
terium without intervention of any other or- 
ganism, as is shown by the infectivity of pure 
DNA. (Although transformation involves 



310 



( HAPTER 23 



MASTER 
PLATE 



T L B 



Contains 



-r -r -r 

♦ TLB, 



REPLICAS 



^ o o o 



- + + 
T L B, 



+ - + + + - 

TLB, TLB, 



B Pa C 



^ o o o 



Medium 
Contains 



t _ -r -i 

| B Pa C 



B Pa C 



+ - + 
B Pa C 



B Pa C 



figure 23—4. Use of replica-plating {shown diagrammatically) to detect spontaneous 
mutations in E. coli. Replica I detects one mutant to T . replica 3 detects one mutant 
to B, . and replica 2' detects one mutant to Pa + . 



the genetic material of two different cells, 
it is not a typical sexual process, since trans- 
formation does not depend upon contact 
between donor and recipient cells.) 

2. The integration process leading to ge- 
netic recombination requires the presence of 
only a portion of the entire genome of the 
donor cell. (Integration results in a small 
segment of the penetrant donor DNA replac- 
ing a small homologous segment of the host 
genome.) 

At this point it does not seem unreason- 
able to hypothesize that any homologous 
DNA penetrating a bacterial cell can inte- 
grate by the same mechanism involved in 
transformation. Of course there may be 
other means of introducing DNA into a 
cell and, consequently, experiments are now 
designed to test whether or not DNA passes 
from one bacterium to another when two 
are in contact. 

One such experiment 2 starts with a proto- 
trophic strain (K12) of E. coli treated with 

-' Based upon work of J. Lederberg and E. L. 
latum ( 1946). 



a mutagen (like X rays or ultraviolet light) 
to obtain single auxotrophic mutants which 
require different nutritional supplements in 
order to grow. The mutagenic treatment is 
repeated — first on the single and then on 
the double mutant auxotrophs — eventually 
obtaining two lines which differ from each 
other by three nutritional mutants, all six 
mutants having arisen independently. One 
triple mutant strain is auxotrophic for threo- 
nine (T~), leucine (L ). and thiamin 
( #, ) ; the other triple mutant is auxotrophic 
for biotin (B~), phenylalanine (Pa), and 
cystine (C~). The genotypes of these two 
lines can be given, respectively, as 



and 



T-L-B 1 B + Pa+C+ 



T + L + B, + BPaC-. 



Of course, the given gene sequence may be 
different in the linkage map. 

The pure lines are grown separately on 
complete liquid culture medium, that is, one 
which contains all nutrients required for 
growth and reproduction. To form a bac- 



Bacterial Mutation and Conjugation 



311 



terial lawn, about 10 s bacteria from one line 
are plated onto agar containing complete 
medium. Then three replica plates are made 
for the TLB X - line (Figure 23-4), each 
plate contains complete medium deficient in 
a different single nutrient (T, L, and B t , 
respectively). Occasionally, a replica shows 
a clone that is able to grow because a pre- 
adaptive mutant produces prototrophy for 
the nutrient missing from the medium. 
However, such clonal growth is not found 
in the corresponding position on all three 
replicas (or even on two) with greater than 
chance frequency. The same results are ob- 
tained when an equal number of bacteria of 
the B~Pa~C~ line are plated on complete 
medium and tested on appropriate replicas. 
We may conclude, therefore, that on rela- 
tively rare occasions mutants to prototrophy 
for one nutrient do occur singly, but double 
or triple mutants do not occur with detect- 
able frequency. 

In another test, the preceding experiment 
is repeated exactly, with the exception that 
the same numbers of the two triply-mutant 
strains are mixed in the liquid medium before 
being plated on agar containing complete 
medium. In this case (Figure 23-5), six 
replicas are made with medium which is 
complete except that three lack B, Pa, and 
C in addition to lacking T, L, or B x ; the 
other three lack T, L, and B, and also B, 
Pa, or C. Individuals of the T~L~B X ~ 
strain cannot grow on the first three replicas 
mentioned because a single required nutrient 
is missing; they cannot grow in the last three 
because all three required nutrients are miss- 
ing. Individuals of the B~Pa~C~ strain 
cannot grow on the first three replicas be- 
cause all three required nutrients are miss- 
ing; they cannot grow on the last three be- 
cause one of the three is absent. If the 
master plate contains a mutant preadaptive 
to nutritional independence for one of the 
nutritionally dependent loci, in only one of 
the six replicas will the mutant form a colony. 



For example, if a T+ mutant occurs among 
the individuals of the T~L-Bi~ strain on 
the master plate, a colony will grow only 
on the replica lacking B, Pa, C, and T. 
Actually, about a hundred different positions 
on the master plate show growth on the 
replicas. This number is very much larger 
than that found in the two groups of three 
replicas made after plating the two lines 
separately. In the present case, some posi- 
tions show growth only on one of the six 



+++ ++ + ___ 

TLBBPaC T L B, B Pa C 



\ / 

Mixed 



MASTER PLATE 




++++++ ^ 
T L B,B Pa C •Medium Contains 

SIX REPLICAS 

2 3 



.9 .<? P 

T L + B,B Pa C T L B,B Pa C T L B,B Pa C 



o.. O. Q.. 

T L B,B Pa + C + T L B,B Pa C T L B,B Pa C 



figure 23-5. Replica-plating (shown diagram- 
matically ) to detect genetic recombination in 
E. coli. A completely prototrophic recombi- 
nant is found at 12 o'clock in all replicas. A 
recombinant for both Pa and C is found at 
3 o'clock on replicas 5 and 6. Replica 1 has 
a clone growing at 9 o'clock which may be 
due either to recombination or to mutation to 
T + . 



312 



CHAPTER 23 



replicas. Many positions, however, show 
growth on two replicas suggesting that 

these clones must have gained nutritional 
independence at two loci. Finally, at 

many positions growth occurs on all six 
replicas, each position of growth represent- 
ing the occurrence of a complete prototroph 
i I I. Bj B l\i C ). A study of these 
clones on the replicas and on the master 
plate shows that the changes involved 
are transmissible and preadaptive. When 
tested, such clones prove to be pure; that 
is, the nutritional independence gained is not 
attributable to any type of physical associa- 
tion between two or more different auxo- 
trophs. Since the findings that large num- 
bers of clones grow on the replicas and 
many are either complete prototrophs or 
auxotrophs for only one nutrient cannot be 
due to spontaneous mutation, they must be 
attributed to some type of genetic recom- 
bination. 

Could this genetic recombination be the 
result of transformation? Recall that most 
transformations involve single loci, and that 
the frequency of double transformations is 
much lower than the frequencies of single 
transformations even when the loci are ap- 
parently very close together, while double 
recombinants are common in the present ex- 
periment. In fact, certain recombinations for 
two loci (for example T+L+BrB+Pa+C+) 
occur more frequently than do recombina- 
tions for these loci singly 



and 



T+L-BrB + Pa + C+ 



T-L + B 1 -B + Pa + C + . 



Transformation explains even less readily 
the large number of triple recombinants — 
the complete prototrophs. Nevertheless, 
some additional experiments are performed 
to test the transformation explanation. The 
number of prototrophs obtained by recom- 
bination is found to be uninfluenced by addi- 
tion of DNase to the medium in which the 



bacteria are mixed or on which the bacteria 
arc plated. No transformation occurs when 
one culture of E. coli is exposed to filtrates 
or autolysates of another. Therefore, the 
genetic recombination detected in E. coli is 
not due to transformation. 

In still another experiment, the base of 
a U-tube is filled with sintered glass to sepa- 
rate the arms; broth is then added, and each 
of the two bacterial strains placed in differ- 
ent arms. Since the sintered glass acts only 
as a bacterial filter, the nutrient medium, 
soluble substances, and small particles (in- 
cluding viruses) can be shuttled back and 
forth. Yet, when platings are subsequently 
made from either arm, no recombinants are 
found. Consequently, the recombinations 
observed are not dependent upon a virus. 
Moreover, after plating a mixture of three 
different mutant lines of K12, we can ex- 
plain all the different recombinants obtained 
as resulting from recombination between any 
two lines; no individuals obtained are re- 
combinant with respect to markers in all 
three lines. Apparently, this type of genetic- 
recombination depends upon conjugation; 
that is, it involves actual cell-to-cell contact 
of bacteria in pairs and is, therefore, a sex- 
ual process. 

The frequency of sexual recombination in 
the first experiment discussed is only about 
one per million cells (100 clones of recom- 
binants per 100 million bacteria placed on 
the master plate). At this point in the in- 
vestigation, the rarity of such events makes 
it fruitless to search for microscopic evidence 
of bacterial mating. (The importance of a 
new phenomenon should not be judged by 
the frequency with which it occurs in experi- 
ments first detecting it. Recall, for example, 
that the quantity of DNA first synthesized 
in vitro was infinitesimal compared with the 
amount synthesized in later work; and the 
rate of transformation observed initially was 
very much smaller than the 10-25% rate 
currently obtained with modified techniques.) 



Bacterial Mutation and Conjugation 



313 



+ + r 



+ + - - s 



BMPTV, xBMPTV, P 



--++s + + — r 
BMPTV, xBMPTV, 



+ + + + 
Prototrophs BMPT 




86% 



V, 



14% 



+ + + + 
Prototrophs BMPT 




79% 
s 
V, 



21% 
r 
V, 



figure 23—6. Genetic recombinations obtained in reversed crosses in- 
volving unselected markers. 



The type of nutrient medium used in the 
experiments described permitted the detec- 
tion of certain recombinants and not others. 
In the first recombination experiment, only 
selective markers — only those alleles which 
provided nutritional independence— were 
selected for detection. Thus, though the 
prototroph T + L+B 1 + B+Pa+C + was detect- 
able, it was not possible to test for the oc- 
currence of the complementary polyauxo- 
troph J-L^B x ~B~Pa'C~ . Since no test 
was made for the multiple auxotroph, one 
could doubt its occurrence. It is entirely 
reasonable that the immediate product of 
mating is a zygote containing part or all of 
the genotypes of both of the parental cells. 
Although it is assumed that integration can 
take place as it does in transformation (if 
DNA passes from one conjugant to another) , 
no evidence that it does has been presented 
so far. In other words, the possibility re- 
mains, at least with respect to the genes 
showing recombination, that the recombinant 
produced by conjugation may be partially or 
completely diploid. 

Although it is not feasible to find zygotes 
microscopically, one can look for them ge- 
netically. After mating, certain clones be- 
have as though they are mosaic for a num- 
ber of markers. When a single cell from 
such clones is isolated, grown, and tested, 



it is found 3 that some of its progeny possess 
the original genotype of either parent and 
that others are recombinants. Clearly, the 
single-cell isolates were derived from more 
or less persistent heterozygotes — individuals 
diploid for various markers. Thus, the hap- 
loid recombinants within a clone offer un- 
ambiguous proof that they were derived from 
a true zygote. Haploid recombinants are 
also called segregants, although this term 
does not imply that they were the result of 
a meiotic process. One concludes, there- 
fore, that the (partially or completely) dip- 
loid zygote produced after bacterial conjuga- 
tion usually has a temporary existence — just 
as in the case of transformation — which 
terminates in the production of haploid 
progeny. In other words, all the genes in 
a nuclear body of E. coli are usually haploid. 
At the same time that certain genetic 
markers are used selectively — to detect re- 
combinant prototrophs, for example — other 
markers may be present that are not subject 
to selection. Such genes are called unse- 
lected markers. E. coli is available in two 
genetic forms: V{ is resistant to infection 
by the bacterial viruses Tj and T, ; ; IV is 
sensitive to infection by these viruses. In- 
fection is followed by death of the bacteria. 

3 By J. Lederberg and M. Zelle. 



:U4 



( II \l' MR 23 



Using this marker and the auxotrophic mu- 
tants P (proline-requiring) and M (meth- 
ionine-requiring), we can make the cross * 

B M /' / I , h\ B \i I' I I', and 
select for the prototrophs B M /' / 
Since the medium employed is virus-free, 

the I , locus is used as an unselected marker. 

The number o\ prototrophs thus obtained 
are subsequently tested tor sensitivity and 
resistance to virus T,. and 86% are found 
typicall) resistant. IV. and L49S typically 
sensitive, I', (Figure 23-6). Note that 
both these alternatives are recombinant rela- 
tive to some of the markers for prototrophy. 

When the reverse cross is made, V { enter- 
ing with the B M + P~T~ parent and V x " 
with B M~P + T+. the percentages of proto- 
trophs sensitive and resistant to T, are found 
to be approximately reversed. In other 
words, the parent that provides P f and T^ 
to the prototroph also contributes the K, 
locus which it contains 79-86% of the time. 
The imbalance in the frequency of resistants 
and sensitives among prototrophs (that is, 
their ratio is not 50% :50% ). and its re- 
versal when the K, markers are reversed in 
the parent cells, provide clear evidence that 
the Vi locus does not integrate and segre- 
gate independently of the other markers 
with which it enters the zygote and which, 
subsequently, are present in the haploid 
prototroph. As a consequence V ', must be 

1 See J. Lederberg (1947). 



linked to /' and / and segregated from these 
loei (or fail to be integrated in the same seg- 
ment with them) onl\ about 20% of the 
time. 

The linkage relationships for auxotrophic 
markers can also be determined, as follows: 
the cross / / B x -B~ by T L B, B is 
made on a complete medium, plated on a 
complete medium, and replicated four times 
on a complete medium without T, or L, or 
Bi, or B. As a result, prototrophic recom- 
binants grow on all four replicas and single 
auxotrophs grow on three, whereas double 
auxotrophs grow on two of the replicas. 
Since prototrophs are found to be more 
frequent than either T L~Bi + B + or 
T L + B : + B • , T and L must be linked. A 
further analysis of the results, for other 
markers and other experiments, reveals that 
all the genetic markers tested in E. coli are 
linked to each other and on a map can be 
arranged in a linear order according to their 
recombination (segregation or integration) 
distances. In all likelihood, this analysis 
indicates, that E. coli has a single ''chromo- 
some. " 

Genetic recombination by the sexual proc- 
ess of conjugation is known to occur in bac- 
teria like Pseudomonas, and — under special 
conditions — in Serratia and Salmonella, as 
well as in Escherichia. Intergeneric conju- 
gation between Escherichia and Salmonella 
has been observed to occur in a mammalian 
host. 



SUMMARY AND CONCLUSIONS 

Bacterial clones are excellent experimental material for the study of the mutation 
process and its rate. Techniques for detecting and isolating mutants are described. 
Mutants occur spontaneously, independent of the factors to which they may he adaptive. 
Genetic recombination occurs in Escherichia coli and other bacteria after the sexual 
process of conjugation. E. coli normally has a haploid nuclear body, in which all 
tested genes belong to a single linear linkage group. 



Bacterial Mutation and Conjugation 



315 



REFERENCES 

Lederberg, J., "Gene Recombination and Linked Segregation in Escherichia Coli," 
Genetics, 32:505-525, 1947. Reprinted in Papers on Bacterial Genetics, Adelberg, 
E. A. (Ed.). Boston: Little, Brown, 1960, pp. 247-267. 

Lederberg, J., "Bacterial Reproduction," Harvey Lect., 53:69-82, 1959. 

Lederberg, L. and Lederberg, E. M., "Replica Plating and Indirect Selection of Bac- 
terial Mutants," J. Bact., 63:399-406, 1952. Reprinted in Papers on Bacterial 
Genetics, Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, pp. 24-31. 

Lederberg, J., and Tatum, E. L., "Gene Recombination in Escherichia Coli," Nature. 
London, 158-558, 1946. Reprinted in Classic Papers in Genetics, Peters, J. A. 
(Ed.), Englewood Cliffs, N.J.: Prentice-Hall, 1959, pp. 192-194. 

QUESTIONS FOR DISCUSSION 

23.1. In previous chapters, has any evidence been presented that spontaneous muta- 
tions are preadaptive? If so, state where. 

23.2. Suppose one thousand streptomycin-free test tubes are each inoculated with 
one bacterium from a streptomycin-sensitive clone, and growth is permitted 
until about one billion bacteria are present in each tube. When the contents of 
each tube are plated on streptomycin-containing agar, what kind of result 
would you expect if the mutations to streptomycin-resistance were postadaptive 
in origin? Preadaptive in origin? From the expected results can you distinguish 
between these two alternatives? How? 



Joshua Lederberg and Esther Marilyn Lederberg about 1951 . {Courtesy 
of The Long Island Biological Association.) 




316 CHAPTER 23 

23.3. From our discussion ol drug-resistant mutants, is it valid to conclude that no 
drugs arc mutagenic.' Explain. 

23.4. Criticize an attempt to prove the occurrence oi genetic recombination by con- 
jugation in which two bacterial strains, each singly auxotrophic lor different 
nutrients, are mixed and subsequently tested for prototrophs. 

23.5. \\h\ is it futile tii search c\tologicall\ for bacteria in the process of conjugation? 

23.6. Differentiate between and give an example of a selective and an unselccted 
marker: a singl) auxotrophic and a prototrophic bacterium. 

23.7. In this chapter, what is the evidence that the two auxotrophs which produced 
prototrophic progeny did so by conjugation rather than by mutation, genetic 
transformation, or viral infection? 

23.8. In what ways does genetic transformation differ from conjugation? 

23.9. Invent suitable genotypes for parents, a zygote, and its clonal progeny, and 
prove the existence of the zygote from the genotypes of the members of the 
clone it produces. 

23.10. Ignoring the centromere, draw all the different ways you can represent a chro- 
mosome whose recombination map is linear. 

23.11. Do you suppose that the discovery of sexuality in bacteria has important im- 
plications for the practice of medicine? Why? 



Chapter 24 

THE EPISOME F 



A' 



re the members of a pair of 
conjugating bacteria equiva- 
lent in that DNA from one 
can go into the other, each bacterium ca- 
pable of acting either as donor or recipient? 
Let us take a look at two streptomycin- 
sensitive and auxotrophically different lines 
able to conjugate with each other to pro- 
duce recombinant progeny. If both lines 
are exposed to streptomycin before — but 
not after — being mixed and plated, none of 
the pretreated individuals can divide. In 
fact, all eventually die, and no recombinant 
clones are formed. When one of the two 
parental lines is pretreated with streptomy- 
cin, again no recombinants are detected. 
But when the other parental line is pre- 
treated, prototrophic recombinants do oc- 
cur. 1 This finding demonstrates that the 
two parents are not equivalent. The parent 
giving no recombinants when pretreated 
acts as the DNA-receiving cell which nor- 
mally would become the zygote after con- 
jugation. When this parent is killed by 
streptomycin, it is impossible to obtain re- 
combinant clones. The other type of par- 
ent must always serve as DNA donor in 
conjugation. After acting as donor, the 
death of this parent has no effect upon the 
zygote and subsequent recombination. The 
parent acting as genetic donor is called F+ 
(for "fertility'"); the parent acting as genetic 
recipient is called F~ . These types serve, 
so to speak, male and female functions, re- 
spectively. In bacterial conjugation, there- 

!See W. Hayes (1953). 
317 



fore, the genetic transfer is a one-way 
process. 

In the discussion on pages 310 through 
312, the original wild-type strain of E. coli 
K12 was F + , and an F - variant must have 
arisen while one of the triply auxotrophic 
lines was being prepared. F+ by F - crosses 
are fertile (show recombination); F~ by 
F - crosses are sterile (show no recombina- 
tion). F^ by F+ crosses can be fertile be- 
cause F+ cells can, on occasion, sponta- 
neously change to F" or because F+ cells — 
acting as F- phenocopies — can occasionally 
behave like F~ phenotypically, despite be- 
ing genetically F+. If one F+ cell is placed 
in a culture of F~ cells, all the F~ cells will 
be rapidly converted to F+ type. F~ cells 
converted to F+ produce F+ progeny. The 
rapidity of change from F~ to F+ is such 
that the causative agent of F+ must mul- 
tiply at least twice as fast as the typical cell 
(and, therefore, twice as fast as chromo- 
somal DNA). Consequently, in E. coli, F + 
male sexuality is an infectious phenomenon 
due to a factor or particle called F. 
Several properties of F are known: 

1. F is transferred from male to female 
only upon contact and cannot be isolated as 
a cell-free particle retaining sex conversion 
potency. (Accordingly, it does not give evi- 
dence of being a typical virus.) 

2. Only one particle of F is transferred 
per mating. 

3. Matings that transfer F are more fre- 
quent but less stable than matings involving 
chromosomal transfer. In fact, the transient 
conjugations which transfer F do not trans- 
fer known chromosomal markers. 

4. Exposure of F 4 individuals to the dye 
acridine orange inhibits the replication of F 
so that F~ cells appear among the progeny. 
Acridine dyes also inhibit the synthesis of 
chromosomal DNA although not as com- 
pletely as they inhibit F factor replication. 
Thus, acridine "curing" is really a differen- 



318 



i HAPTER 24 



tial phenomenon. In stable F cultures, F 
multiplication proceeds al exact!) the same 
rate as does chromosomal multiplication. 
This means that there is precise regulation 

of the number of F particles per chromo- 
some and it is onl\ when this regulation is 
inoperative — when a population of F cells 
is being converted to F b\ a tew F+ cells, 
lor example — that one sees the explosive 
multiplication of F. 

These evidences are sufficient to conclude 
that / is an extrachromosomal panicle. F 
not only makes a cell a potentially fertile 
male — that is. potentially capable of acting 
as chromosomal donor — but it affects the 
cell harboring it in several ways: 

1. It changes the surface of a male cell 
so that on contact, it can recognize and react 
with a female cell. 

2. It must be the cause of some kind of 
bridge between the male and female cell 
through which F or the chromosome is 
passed to the F~ cell. 

3. It must also cause the formation of a 
receptor at the surface of the cell for viruses 
that attack males only.- 

4. Finally, F has the ability to initiate, at 
a low frequency, the transfer of random 
chromosomal markers during F + by F~ 
crosses. This phenomenon has been termed 
Lfr (low frequency of recombination). 

So far we have found E. coli to have two 
mating types, F~ and F . Another mating 
type can arise from F+ cells. This type 
produces a rather /ngh frequency of recom- 
bination of chromosomal genes; hence its 
name. Hfr. Since the fertility of Hfr cells 
is unaffected by pretreatment with strepto- 
mycin. Hfr cells are donors. These cells 
can mate with F~ cells, and — although fer- 
tility is low — with cells in F cultures 
(which have either changed spontaneously 
to F - or act as F phenocopies). Crosses 

-According to T. Loeb and N. D. Zinder (1961). 



oi Hfr with F ; produce 100 to 20,000 times 
as main recombinants as F+ by F crosses. 
Since the progenj o\ Hfr by F arc typically 
F~ and rarely Hfr. Hfr does not usuall) 
carry infective F particles. Hfr can. how- 
ever, revert to F strains apparently show- 
ing all the characteristics of F + , including 
infective F. Since Hfr can only come from 
F+ and can only revert to F . it is con- 
cluded that F must be retained in masked 
or bound form in Hfr strains. When F is 
present in this masked condition, the replica- 
tion of any extrachromosomally located F 
capable of infection is suppressed. 

Using an Hfr strain, the cytological search 
for conjugating pairs is successful. Figure 
24-1 is an electron micrograph showing 
conjugation between an F^ cell and an Hfr 
cell. The Hfr cell has ultraviolet-killed bac- 




figure 24-1. Conjugation in E. coli. {Cour- 
tesy of T. F. Anderson.) 



The Episome F 



319 



terial virus particles (tadpole-shaped ob- 
jects) adsorbed on its surface; the F - cell 
does not, since it is genetically resistant to 
this virus. The cytoplasmic bridge between 
the conjugants is clearly shown. When ex- 
conjugants of such visibly marked pairs of 
Hfr and F~ cells are isolated by micro- 
manipulation and are cultured, only the 
clones from the F~ partner yield recom- 
binants. Note, in passing, that these find- 
ings again demonstrate how genetics and 
cytology have aided each other's advance- 
ment. 

In studies of Hfr by F~ crosses, it is com- 
monly found that most of the unselected 
markers in recombinant progeny are those 
derived from the F~ parent. This result 
may be explained either by the transfer of 
the entire genome of the male to the female 
followed by the integration of only a portion 
of it, or by the transfer of only a portion of 
the male genome and its integration in toto, 
or by a combination of the two. Experi- 
ments have been designed to test one or 
more of these possible explanations/' 

Particular strains of Hfr and F~, both 
marked with suitable genetic factors, are 
grown separately and then mixed in the pro- 
portion of 1 : 20, respectively, to assure rapid 
contact of all Hfr with F~ cells. At various 
time intervals, ending roughly two hours 
after mixing, samples are withdrawn and 
subjected to a strong shearing force in a 
Waring blendor. This treatment is a very 
efficient means of separating bacteria in the 
act of conjugation and does not affect either 
the viability of the bacteria, their ability to 
undergo recombination, or the expression of 
the various chromosomal genotypes under 
test. Once separated, the bacteria are plated 
and scored for male markers which have 
integrated. As expected, zero minutes after 
mixing no recombinants for male markers 

:; The following discussion is based principally 
upon work of E. L. Wollman and F. Jacob (see 
references at the end of this chapter). 



MINUTES RECOMBINANTS HAVING Hfr MARKERS 

None 

8 T 
8'A T, L 

9 T, L, Az 

11 T, L, Az, T, 

18 T, L, Az, T, , Lac 

25 T, L, Az, T, , Lac, Gal 

figure 24-2. Recombinants obtained when 
conjugation is artificially interrupted at various 
times after mixing F~ and Hfr strains. The 
Hfr strain has markers for T, L, Az, 7\, Lac. 
Gal. (After W . Hayes.) 



are obtained; to obtain recombinants for all 
the male markers, conjugation must last for 
about 107 minutes. The time at which dif- 
ferent male markers enter the female cell, 
however, varies widely within this time in- 
terval (Figure 24-2). For example, T and 
L markers of the Hfr do not enter F~ until 
after about 8.5 minutes of conjugation, 
whereas the Gal marker (for galactose) re- 
quires about 25 minutes of conjugation be- 
fore it is transferred. T and L are known 
to be close together and widely separated 
from Gal in the recombination map of F + 
(p. 314); consequently, a definite relation- 
ship exists between time of transfer from 
Hfr to F" and the location of the marker 
on the Hfr chromosome. 

Had different portions of the Hfr chro- 
mosome entered the F - cell at random 
times, the above results would not have been 
obtained. It can be concluded, therefore, 
that the Hfr chromosome is transferred in a 
specific manner: one particular end of the 
DNA string usually enters the F - cell first, 
since the loci that transfer do so in a regular 
linear procession (Figure 24-2). Other ex- 



H20 



CHAPTER 24 



periments reveal that energy is required for 
the transfer process, and that the entrance 
rate is uniform from the first part of the 
chromosome to be transferred. O ( represent- 
ing the "origin"), up to and including the 
locus of Lac (for lactose ). 

Whether or not they receive or lose a seg- 
ment of an artificially ruptured chromosome, 
both the female and male cells are able to 
sun ive. Chromosome rupture may also oc- 
cur spontaneously. Because of such break- 
age, the spontaneous transfer of the Hfr 
chromosome is usually partial. Conse- 
quently, a piece of chromosome, called a 
merogenote, of variable size is sent into the 
recipient cell. As a resut. the zygote of an 
Hfr cross is a partial diploid — a merozygote 
— produced by a process of partial genetic 
exchange, meromixis. These three new 
terms are applicable also in transformation. 

The frequency of recombination is rela- 
tively low for all chromosomal genes in an 
F^ strain. In Hfr, however, this frequency 
is high for those markers nearest O, although 
it decreases as the distance of the markers 
from O increases, and is only 0.001-0.01% 
for the markers furthest from O. As men- 
tioned previously, only rarely is an offspring 
of Hfr by F- itself Hfr. An Hfr offspring 
arises only when the marker furthest from 
O — the terminal marker — has also been 
transferred. This fact leads us to believe 
that the locus responsible for Hfr is located 
in the chromosome. Moreover, since no 
chromosomal locus is found transferred after 
the Hfr locus, we can conclude that Hfr is 
always located at the terminus of a chromo- 
some being transferred. 

Several Hfr strains derived from F^ cul- 
tures show a very high frequency of recom- 
bination. 4 In these strains, recombination 
rates for the markers furthest from O occur 
with a frequency of one to two per cent — 
a rate at least one hundred times that found 

* See A. L. Taylor and E. A. Adelberg ( 1960). 



SELECTED 








MARKER 




STRAIN 






AB-311 


AB-312 


AB-313 


+ o 

his 42 


2.5 


.^ 


+ 
gal 


12 


4 


— 


+ 
pro 


— 


8 


— 


+ 
met 


4 


22 


— 


+ „ ° 
mtl t 3.7 1 25 49 


xyl 


2.8 


26 


43 


+ 
mal 


1.5 


40 


32 


ade 


O 
- - f 15 


<ry + 


— 


— 


6 


+ 
arq 




__ 


0.3 



figure 24-3. Recombination percentages for 
certain Hfr strains. = point of origin; — = 
untested. (After A. L. Taylor and E. A. Adel- 
berg, 1960. See References.) 



in other Hfr strains. Even so, less than 
about one per cent of the progeny from mat- 
ing these Hfr with F are Hfr. By artificial 
rupture experiments, the sequence of certain 
marker genes can be determined for each 
of the three independently-arisen Hfr strains 
of this type. 

These sequences are shown in Figure 
24-3 together with the frequency of recom- 
binants per one hundred Hfr cells in the 
mating mixture. 

The results show that the markers held 
in common by the three different Hfr strains 
are in the same sequence. The O point, 
however, is in a different position in each 
case! Accordingly, so is the position of the 



The Episome F 



321 



Hfr locus at the end of the linkage map. 
These results suggest " that: 

1. The linkage group of E. coli is nor- 
mally circular. 

2. The Hfr-causing factor can locate itself 
at various places on the chromosome during 
the genetic event by which Hfr strains are 
formed. 

3. At the time of chromosome transfer. 
the linkage group is open adjacent to the 
point of Hfr attachment so that the Hfr locus 
is at the end opposite the O point. 

Studies of other Hfr strains ' confirm all 
these assumptions, including various posi- 
tions for Hfr — which consequently determine 
different O points and sequences of entry. 

When DNA synthesis occurs in the pres- 
ence of radioactive thymidine, autoradio- 
graphs of replicated DNA strongly suggest 
the presence of a single, circular, double- 
helix chromosome in E. coli. The single 
chromosome may be. in fact, a single DNA 
molecule or it may be composed of a series 
of DNA molecules held end-to-end by non- 
nucleic acid links. There is no definitive 
evidence which allows us to decide between 
these two alternatives. It has been found, 
however, that the chromosome tends to frag- 
ment at certain predetermined points during 
the growth of particular bacterial viruses in 
infected cells. If the chromosome is actually 
composed of an assemblage of DNA mole- 
cules and if breakage does occur at all non- 
nucleic acid links, then the DNA molecules 
would have a molecular weight of 10" or 
contain about 16.000 nucleotide pairs. 
Other evidence suggests that the linkage 
group of Hfr is usually circular, though the 
linkage group transferred during conjugation 
is open at the Hfr locus. 

Since an Hfr male strain always has the 
same chromosomal marker leadine the others 



" Following F. Jacob and E. L. Wollman. 

6 See A. L. Taylor and E. A. Adelberg < 1961). 



T L 




Ser/g|y 



figure 2-! — X. Linear chromosomes of three 
Hfr strains. Arrows show direction of chro- 
mosome penetration during conjugation. \ 
A. L. Taylor and E. A. Adelberg. 1960. See 
References.) 



in transfer. Hfr apparently causes the ring 
chromosome to open at only one of the two 
regions immediately adjacent to it. That 
the entry sequence is different in different 
strains (the chromosome of AB-312 enters 
in the reverse direction from that of AB-3 1 1 
or AB-3 13. as can be seen in Figures 24—3 
and 2-1 — I ) may be due to an inversion of 
Hfr when it locates itself at different chro- 
mosomal positions. 

Since the ring chromosome of E. coli is 
opened when the Hfr chromosome is trans- 
ferred, one might question whether these two 
new ends are able to join in restitution. 
WTien considering whether the two new ends 
rejoin, one must distinguish between what 
goes into the recipient cell and what is left 
behind in the donor cell. There is little 
evidence about how much chromosome is 
left behind — whether, in fact, there is or is 
not a complete chromosome — in the donor 



322 



( IIM'TER 24 



after conjugation. Thus, the question of 

whether the ends of the remnant ean rejoin 
or not in the donor cannot be answered at 
present. It is clear that the donated chro- 
mosome remains linear and open during the 
process o{ conjugation. One ean. however, 
legitimately ask whether the two ends of this 
chromosome or chromosome segment can 
rejoin in the recipient after conjugation has 
terminated. Again, if one believes in synap- 
sis as a requirement for recombination, this 
question ean be asked only with regard to 
those recipients which receive a complete 
donor chromosome. In this case, since link- 
age of the Hfr locus and point of origin is 
rapidk re-established in such cells, the an- 
swer is that ends do rejoin, and quickly. 

The recombination frequencies observed 
after conjugation depend, of course, upon 
both the frequency of a marker's penetra- 
tion and the efficiency with which it is in- 
tegrated. Interrupted-mating experiments 
reveal the sequence of markers, regardless 
of the frequency (greater than zero) with 
which their integration occurs. Once the 
marker sequence is known, integration effi- 
ciency can be studied. If, for example, mat- 
ings are permitted to continue long enough 
so that just about all F^ cells are penetrated 
by the marker under test, the percentage of 
zygotes producing recombinants for that 
marker will indicate the efficiency of integra- 
tion. If 50% of the recipient cells show 
integration of a transferred marker, this locus 
has an integration efficiency of .5. One can 
also test whether recombinants for a given 
locus are recombinants for markers trans- 
ferred earlier. By these and other methods, 
the integration efficiency after penetration 
can be determined for various markers. On 
the average, the integration efficiency is 
about .5 for each marker. Therefore, be- 
cause of differences in penetration, the closer 
a gene is to O, the greater is its overall 
chance for integration. Recall that when 
one strand of a double helix of donor DNA 



is broken b\ DNase activity, incorporation 

of donor DNA into the recipient DNA frac- 
tion is unaffected but transformation rate is 
drasticallj reduced. Accordingly, it is the- 
oretically possible that the decreases in the 
overall chance for integration with distance 
from O may sometimes be associated with 
the breakage of one strand of double-helix 
DNA and not the other. In this event, back- 
bone defects might affect not only chromo- 
somal breakage (by scission of the DNA 
double chain), but also integration (by 
breaking only one strand of the two at any 
given level ) . 

Although the donor DNA rate of penetra- 
tion is approximately constant for the first 
half of the chromosome, it is slower for the 
second half. Since the bacterial chromo- 
some is about 10 T nucleotide pairs long and 
the entire chromosome is transferred in about 
two hours, about 10"' nucleotide pairs (about 
34 fji) are transferred in one minute at 37° C. 
This transfer rate is slower at lower tem- 
peratures. Taking into account variations 
in rate of penetration and integration effi- 
ciency, it is possible to construct, from in- 
terruption experiments using Hfr males, a 
genetic map of E. coli markers whose rela- 
tive distances are expressed in minutes. 
Such a map is shown in Figure 24-5. 

It was mentioned that F is transferred 
from F to F - even when known chromo- 
somal markers are not. When an inter- 
rupted-mating experiment is performed to 
determine the time when the F particle in 
F+ males is transferred, it is found that F 
is first transferred about five minutes after 
mixing F f and F or several minutes earlier 
than any known marker in the chromosome 
is transferred. We, therefore, have addi- 
tional evidence that F is extrachromosomal 
in nature. 

As already mentioned, Hfr strains are al- 
ways derived from F+ strains. It was also 
noted that Hfr strains can revert to F + , 
indicating that Hfr harbors a latent F par- 



The Episome F 



323 



ticlc. Since the fertility of Hfr is unaffected 
by exposure to acridine orange, the latent 
F particle is probably not located extrachro- 
mosomally. This contention is supported 
by the fact that maleness is not infective; 
that is, maleness is not transmitted to F~ 
cells after short mating intervals with Hfr 
males. Consequently, the latent F particle 
in Hfr must be located chromosomally. 
Since, at this point, F is the only known 
factor essential for maleness, the chromo- 
somal locus assigned to the Hfr must be 
that of chromosomal F. Once F enters the 
chromosome, as mentioned, replication of 
any remaining cytoplasmic F particles nor- 
mally is prevented or repressed. 

What happens in those few cells of an F + 
clone which transfer chromosomal material, 
causing the F + clone as a whole to give a 
low frequency of recombination? Suppose 
that for an F+ cell to transfer its chromo- 
some, an F particle must attach to the chro- 
mosome, making it an Hfr chromosome. 
This hypothesis can be tested as follows. 7 
After mixing suitably marked F+ and F~ 
and plating them on a complete medium, 
appropriate replica plates are made to de- 
tect the positions where recombination has 
taken place. A search is then made for Hfr 
strains among the cells on the master plate. 
Although new Hfr strains rarely occur, they 
are found most frequently on the master 
plate at positions where replicas show that 
recombination has taken place. Moreover, 
the Hfr strains discovered often produce a 
high frequency of recombination of the same 
markers that show recombination in the cor- 
responding positions in the replicas. In 
other words, it seems valid to believe that to 
transfer chromosomal material, an F+ indi- 
vidual must first change to an Hfr condition. 
It is the Hfr which produces the recombina- 
tion detected on the replica, its clonal mem- 



7 Based upon work of F. Jacob and E. L. Woll- 
man. 



bers on the master plate yielding the same 
type of Hfr. 

Some evidence exists, however, that not 
all recombinants in F + by F _ crosses result 
from the formation of stable Hfr cells. For 
example, UV radiation enhances the ability 
of an F^ culture to give chromosomal re- 
combinants by 30- to 50-fold but this en- 
hancement is not the result of stable integra- 
tion of F into the chromosome since it only 
persists for one or two generations. Fur- 
thermore, all attempts to isolate stable Hfr 
from other bacteria harboring other types of 
fertility factors have been unsuccessful. Ap- 
parently, some sort of temporary or "abor- 
tive" Hfr state is sometimes — if not always 
— produced in F+ cells which transfer chro- 
mosomal markers. 

Since F+ males rarely become Hfr, we 
conclude that extrachromosomal F has a low 
affinity for t/ie chromosome; and, since F+ 
males produce Hfr strains having any one of 
various O points, we conclude that F has no 
preferential site of attachment. F is com- 
posed of DNA (F transferred to Serratia 
gives a satellite DNA band in CsCl not 
found in Serratia alone) and contains 
roughly 2.5 times 10 5 base pairs, or about 
3.7% of the amount present in a nuclear 
body. Although F has a considerable num- 
ber of sites at which it can integrate, this 
number is by no means unlimited in view of 
the repeated isolation of strains with Hfr 
mapping in the same region. In any case, 
if one accepts the idea that naturally-occur- 
ring F possesses regions which are homol- 
ogous to certain chromosomal regions and 
that integration is in fact the result of a 
recombinational event between F and the 
chromosome, then it is reasonable that F, 
because of its size, could carry genetic re- 
gions homologous to only a small fraction 
of all chromosomal sites. 

The bacterial chromosome always repli- 
cates in a polarized manner, starting at one 
point and continuing to the end. Although 



324 



( ii\i'i i i< 24 



Q < CD U 



roa 







^K^i 



o o 



FIGURE 24-5. 

Genetic map of E. coli, 

drawn to scale. A key to 

the genetic symbols appears 

in the accompanying chart. The map 

is graduated in one-minute intervals 

(89 minutes total). Markers enclosed 

in parentheses are only approximately 

mapped; the exact sequence of markers 

in crowded regions is not always known. 

(Courtesy of A. L. Taylor and 

M. S. Thoman, 1964. See References.) 



The Episome F 



325 



List of Chromosomal Markers of E. coli 



Key to genetic symbols * 



Activity affected 



araD 


arabinose 


araA 


arabinose 


araB 


arabinose 


araC 


arabinose 


argB 


arginine 


argC 


arginine 


argH 


arginine 


argG 


arginine 


argA 


arginine 


argD 


arginine 


argE 


arginine 


argF 


arginine 


aroA,B,C 


aromatic amino acids and vitamins 


aroD 


aromatic amino acids and vitamins 


azi 


azide 


bio 


biotin 


cysA 


cysteine 


cysB 


cysteine 


cysC 


cysteine 


dapA 


diaminopimelic acid 


dapB 


diaminopimelic acid 


dap + horn 


diaminopimelic acid + homoserine 


Dsd 


D-serine 


fla 


flagella 


galA 


galactose 


galB 


galactose 


galD 


galactose 


galC 


galactose 


gua 


guanine 


H 


H antigen 


his 


histidine 


He 


isoleucine 


ilvA 


isoleucine + valine 


ilvB 


isoleucine + valine 


UvC 


isoleucine + valine 


ind 


indole 


\ 


prophage \ 


lacY 


lactose 


lac Z 


lactose 


lacO 


lactose 


leu 


leucine 


Ion 


long form 


lys 


lysine 


lys + met 


lysine + methionine 


\rec,malA 


\receptor and maltose 


malB 


maltose 


met A 


methionine 


metB 


methionine 



L-ribulose 5-phosphate 4-epimerase 

L-arabinose isomerase 

L-ribulokinase 

unknown 

N-acetylglutamate synthetase 

N-acetyl-7-glutamokinase 

N-acetylglutamic-7-semialdehyde dehydrogenase 

acetylornithine-5-transaminase 

acetylornithinase 

ornithine transcarbamylase 

argininosuccinic acid synthetase 

argininosuccinase 

shikimic acid to 3-enolpyruvylshikimate-5-phosphate 

biosynthesis of shikimic acid 

resistance or sensitivity to sodium azide 

unknown 

3'-phosphoadenosine 5'-phosphosulfate to sulfide t 

sulfate to sulfide; four known enzymes t 

dihydrodipicolinic acid synthetase 

N-succinyl-diaminopimelic acid deacylase 

aspartic semialdehyde dehydrogenase 

D-serine deaminase 

galactokinase 

galactose 1 -phosphate uridyl transferase 
uridinediphosphogalactose 4-epimerase 
operator mutants 

flagellar antigen 

ten known enzymes and an operator t 

threonine deaminase 

a-hydroxy £-keto acid reductoisomerase t 

a,/3-dihydroxyisovaleric dehydrase t 

transaminase B 

tryptophanase 

galactoside permease 
/3-galactosidase 
operator mutants 

three known enzymes and an operator t 
filament formation and radiation sensitivity 
diaminopimelic acid decarboxylase 
unknown 

maltose permease and resistance to phage \ 
probably amylomaltase 

synthesis of the succinic ester of homoserine t 
succinic ester of homoserine -\- cysteine to 
cystathionine t 



{Continued on next page) 



320 



( IIM'TI K 24 



l isi 01 Chromosomai Markers of E. coli — continued 



Ke\ to genetic symbols 



Activity affected 



metF 

metE 
mtl 

III IIC 

O 
pan 

pheA.B 
pho 

P il 

pro A 

proB 

proC 

purA 

purB 

purC.E 

purD 
pyrA 
pyrB 
pyrC 

pyrD 
pyrE 
pyrF 

Rarg 

Rgal 

Rl pho, R2 pho 

R try 

RC 

rha 

serA 

serB 

sir 

sue 

T1J5 rec 

Tl rec 

T6, colK rec 

T4 rec 

thi 

tin 

thy 

tryA 

iryB 

tryC 

tryE 

tryD 

tyr 

uvrA 

xyl 



methionine 

methionine or cobalamin 

mannitol 

mucoid 
O antigen 
pantothenic acid 
phenylalanine 

phosphatase 

pili i fimbriae) 

proline 

proline 

proline 

purine 

purine 

purine 

purine 

uracil + arginine 

uracil 

uracil 

uracil 

uracil 

uracil 

repressor 

repressor 

repressor 

repressor 

RNA control 

rhamnose 

serine 

serine 

streptomycin 

succinic acid 

phage receptor site 

phage receptor site 

phage and colicine receptor site 

phage receptor site 

thiamine 

threonine 

thymine 

tryptophan 

tryptophan 

tryptophan 

tryptophan 

tryptophan 

tyrosine 

ultraviolet radiation 

xylose 



prohahly 5,10-mcthylene tetrahydrofolate reductase 

unknown 

probably mannitol dehydrogenase 

regulation of capsular polysaccharide synthesis 

somatic antigen 



alkaline phosphatase 



supports syntrophic growth of proA mutants 

supports syntrophic growth of proA mutants 

adenylosuccinic synthetase 

adenylosuccinase 

5-aminoimidazole ribotide(AIR ) to 5-aminoimidazole 

4-(N-succinocarboxamide) ribotide 

biosynthesis of AIR 

carbamate kinase 

aspartate transcarbamylase 

dihydroorotase 

dihydroorotic acid dehydrogenase 

orotidylic acid pyrophosphorylase 

orotidylic acid decarboxylase 

arginine repressor 

galactose repressor 

alkaline phosphatase repressor 

tryptophan repressor 

regulation of RNA synthesis 

utilization of D-rhamnose 

3-phosphoglycerate dehydrogenase 

phosphoserine phosphatase 

resistance, sensitivity, or dependence on streptomycin 

resistance to phages Tl and T5 
resistance to phage Tl 
resistance to phage T6 and colicine K 
resistance to phage T4 



thymidylate synthetase 

tryptophan synthetase, A protein 

tryptophan synthetase, B protein 

indole 3-glycerolphosphate synthetase 

anthranilic acid to anthranilic deoxyribulotidc 

3-enolpyruvylshikimate 5-phosphate to anthranilic acid 

reactivation of UV-induced lesions in DNA 
utilization of D-xylose 



* Established systems of genetic nomenclature are retained wherever possible, except that capital letters be- 
ginning with the letter A are arbitrarily assigned to functionally related gene loci which do not conform to 
the system of bacterial genetic nomenclature proposed by M. Demerec (1963). 

t Denotes enzymes controlled by the homologous gene loci of Salmonella typhimurium. 



The Episome F 



327 



the starting point of replication in F~ is 
possibly random, some evidence indicates 
that replication starts at the F-containing 
end in Hfr strains/ Chromosome transfer 
in Hfr lines may require initiation of a new 
replication of the chromosome." 

The chemical composition, genetic alter- 
natives, 10 capacity for autonomous self-du- 
plication, and "suicide" due to incorporated 
P :i - after transfer to F~, demonstrate that 
the male fertility factor, F, of E. coli must 
be composed of genetic material. Since F 
is probably neither lytic nor otherwise rap- 
idly lethal to F~ cells, and since F has a 
stable relationship with the F+ cell, it can 
be considered a "normal" cellular compo- 
nent, when present. Because it can func- 
tion and reproduce in an autonomous man- 
ner when located extrachromosomally, F 
furnishes the first example so far presented 
of "normal," extrachromosomal genetic ma- 
terial. When extrachromosomal F is lost — 
either spontaneously or after treatment with 
acridine orange — it represents genetic ma- 
terial not conserved for future generations. 
Since F is genetic material, its transmission 

8 See J. Cairns (1964), T. Nagata (1964), and 
N. Sueoka and H. Yoshikawa (1964). 

9 See J. Roeser and W. A. Konetzka (1964). 

10 See E. A. Adelberg and S. N. Burns (1960). 



from F+ to F - is an example of genetic 
recombination. 

F also has the ability to assume a regular 
locus on a chromosome. When stably at- 
tached to or integrated in the chromosome, 
F functions and replicates just as any other 
ordinary chromosomal locus. In regular 
vegetative reproduction, chromosomal F is 
transmitted to all progeny; that is, F is con- 
served. In conjugation, however, chromo- 
somal F may not be conserved, for inte- 
grated F is not transmitted to the zygote 
with appreciable frequency except in the 
case of certain Hfr strains, and — even when 
transmitted — may fail to be integrated. In 
such cases, the nonconservation of F in F~ 
cells is no different from the nonconserva- 
tion of other chromosomal loci. 

The only type of genetic elements dis- 
cussed in detail prior to this chapter were 
those restricted to the chromosome. To 
these can now be added the male fertility 
factor, F, which may or may not be present 
in the cell, and, when present, can be either 
autonomous extrachromosomally or inte- 
grated in the chromosome. Such genes 
which can participate in the cell either as 
extrachromosomal or as chromosomal ele- 
ments are called episomes. 11 

11 See F. Jacob and E. L. Wollman (1958). 



SUMMARY AND CONCLUSIONS 

E. coli has only one female mating type (genetic recipient) — F~ — but two male mating 
types (genetic donors) — F+ and Hfr. Male mating type depends upon the presence 
and location of F. 

F, an episome, is infective when present extrachromosomally. Thus, F+ males 
donate F only. F can assume a chromosomal locus and produce the Hfr male. When 
the Hfr male conjugates, the ring chromosome of E. coli is open near the locus of F. 
The end opposite the F locus proceeds into the recipient conjugant first, in the linear 
transfer of part — or sometimes all — of the opened ring chromosome. Thus, Hfr males 
mobilize and donate chromosomal loci. 



328 



CHAPTF.R 24 




Elie L. Wollman (left) and Francois Jacob (right) in 1961. 



REFERENCES 

Adelberg. E. A., and Burns, S. N., "Genetic Variation in the Sex Factor of Escherichia 
Coli," J. Bact., 79:321-330, 1960. Reprinted in Papers on Bacterial Genetics, 
Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, pp. 353-362. 

Cairns, J., 'The Chromosome of Escherichia coli," Cold Spring Harb. Sympos. Quant. 
Biol., 28:43-46, 1964. 

Clark. A. J., and Adelberg, E. A., "Bacterial Conjugation," pp. 289-319, in Annual 
Review of Microbiology, 16, Palo Alto, Calif.: Annual Reviews, Inc., 1962. 

Hayes. W., "The Mechanism of Genetic Recombination in Escherichia coli," Cold Spring 
Harb. Sympos. Quant. Biol.. 18:75-93, 1953. Reprinted in Papers on Bacterial 
Genetics, Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, pp. 268-299. 

Hayes, W., The Genetics of Bacteria and Their Viruses, New York: J. Wiley & Sons, 
1964. 

Jacob, F., and Wollman, E. L., "Episomes, Added Genetic Elements" (in French), 
C. R. Acad. Sci. (Paris), 247:154-156, 1958. Translated and reprinted in Papers 
on Bacterial Genetics, Adelberg, E. A. (Ed.), Boston: Little, Brown. 1960, pp. 
398-400. 

Jacob, F., and Wollman, E. L., "Genetic and Physical Determination of Chromosomal 
Segments in Escherichia Coli," Sympos. Soc. Exp. Biol., 12:75-92, 1958. Re- 
printed in Papers on Bacterial Genetics, Adelberg, E. A. (Ed.). Boston: Little. 
Brown, 1960, pp. 335-352. 

Jacob, F., and Wollman, E. L., Sexuality and the Genetics of Bacteria, New York: 
Academic Press, 1961. 

Nagata. T.. "The Sequential Replication of E. coli DNA," Cold Spring Harb. Sympos. 
Quant. Biol., 28:55-57, 1964. 



The Episome F 329 

Roeser, J., and Konetzka. W. A., "Chromosome Transfer and the DNA Replication 
Cycle in Escherichia coli," Biochem. Biophys. Res. Commun., 16:326-331, 1964. 

Sueoka, N.. and Yoshikawa. H., "Regulation of Chromosome Replication in Bacillus 
subtilis," Cold Spring Harb. Sympos. Quant. Biol.. 28:47-54, 1964. 

Taylor. A. L., and Adelberg, E. A., "Linkage Analysis with Very High Frequency 
Males of Escherichia Coli," Genetics, 45:1233-1243, 1960. 

Taylor, A. L., and Adelberg, E. A., "Evidence for a Closed Linkage Group in Hfr 
Males of Escherichia coli K-12," Biochem, Biophys. Res. Commun., 5 400-404 
1961. 

Taylor, A. L., and Thoman, M. S., "The Genetic Map of Escherichia coli K-12," 
Genetics, 50:659-677, 1964. 

Wollman, E. L., Jacob, F., and Hayes, W., "Conjugation and Genetic Recombination 
in Escherichia coli K 12." Cold Spring Harb. Sympos. Quant. Biol., 21:141-162, 
1956. Reprinted in Papers on Bacterial Genetics, Adelberg, E. A. (Ed.), Boston: 
Little, Brown, 1960, pp. 300-334. 



QUESTIONS FOR DISCUSSION 

24.1. What events do you suppose occur from the time donor DNA enters a female 
conjugant to the appearance of a segregant haploid for a segment of the donor 
DNA? 

24.2. What is the evidence that F can integrate in a chromosome? That it can do 
the reverse (deintegrate)? 

24.3. What properties are attributable to F when integrated at a chromosomal locus? 

24.4. Does the occurrence of spontaneous ruptures of the donor bacterial chromo- 
some interfere with mapping the linear order of genes via artificial interruptions 
of mating? Explain. 

24.5. Assume (correctly) that the decay of P 32 incorporated into DNA can break the 
E. coli chromosome, and that this decay is temperature-independent. Devise an 
experiment to determine the gene order in this bacterium. 

24.6. What kinds of evidence can you present that a nuclear body of E. coli normally 
contains a single chromosome? A ring chromosome? 

24.7. Give the genotypes of parents and recombinants and the specific culture condi- 
tions you would employ in searching the progeny of F+ by F^ crosses for Hfr 
lines. 

24.8. What relationships exist between episomes and genetic recombination? 

24.9. Transforming DNA is isolated from cultures of Bacillus subtilis in a stationary 
(nonreplicating) phase and in an exponentially-growing and nonsynchronously- 
reproducing phase and tested with respect to eleven genetic markers. Invent 
quantitative results expected if the chromosome replicates in a polarized manner. 



Chapter 25 
TRANSDUCTION 



T! 



(he three preceding chapters 
dealt with recombination of 
genetic material in bacteria 
by mechanisms usually involving relocation 
of only a portion of a bacterial genome. 
In transformation, the presence of a donor 
organism is unnecessary for the entrance of 
donor DNA; in conjugation, however, a seg- 
ment of chromosomal DNA passes from 
donor to recipient bacterium through a cy- 
toplasmic bridge effected by the presence of 
an F particle. The F particle itself under- 
goes recombination not only when it is in- 
fectious but when it enters and leaves the 
bacterial chromosome. 

As already noted (p. 310), it is expected 
that any homologous segment of DNA, no 
matter how introduced into a bacterium, can 
pair with and integrate into a bacterial chro- 
mosome. A third possible mechanism ex- 
ists for introducing homologous DNA into 
bacteria. Bacteriophages or phages 1 arc 
viruses that attack bacteria. After these vi- 
ruses become attached to the bacterial sur- 
face (see Figure 24-1), all or part of the 
phage remaining external to a bacterium can 
be shaken off by the shearing action of a 
blendor. Such treatment does not alter the 
course of the infection, however; that is, the 
virus still produces its characteristic effect 
on the bacterium. We can infer that the 
part of the virus essential for this effect 
actually enters the bacterial protoplasm, 

1 The Greek letter <$> (phi) is used to denote phage. 
330 



whereas the part of the phage that remains 
attached to the bacterial surface is unneces- 
sary. These observations suggest a new 
way by which homologous DNA may en- 
ter a phage-infected cell. The virus might 
carry a segment of DNA derived from a 
previous bacterial host. This piece could 
penetrate the new host at the same time as 
part of the phage docs, the phage's entry 
providing the opening for the bacterial 
DNA. 

With this possibility in mind, consider a 
series of experiments 2 involving the mouse 
typhoid organism. Salmonella typhimurium. 
This bacterium, like its close relative E. coli, 
can also be cultured on a simple nutrient 
medium. A large number of auxotrophic 
strains of Salmonella are available, includ- 
ing one that requires methionine (M~T+ ) 
and another that requires threonine 
(M + T~). When these two strains are 
mixed and plated on a culture medium lack- 
ing both methionine and threonine, proto- 
trophic colonies appear in such large num- 
bers that they cannot be explained entirely 
as the result of mutation. Prototrophs are 
also obtained when a liquid culture of the 
M + T~ strain is centrifuged (to remove 
most of the bacteria), the supernatant liquid 
heated for 20 to 30 minutes (to kill any 
remaining bacteria), and this liquid added 
to the M~T+ strain. This procedure dem- 
onstrates that living M + T donor cells are 
not required to furnish the M+ factor 
needed to establish prototrophy. So here 
the production of prototrophs clearly docs 
not result from conjugation. Moreover, the 
filtrate retains its full M+ capacity after 
treatment with DNase. Accordingly, this 
is not a case of genetic recombination via 
transformation. Since the M+ factor can 
pass through filters that hold back bacteria 
but not viruses, the factor is a "filterable 
agent." The reverse experiment — using fil- 

'-' The following discussion is based upon the work, 
of N. D. Zinder and J. Lederberg (1952). 



Transduction 



331 



trates of the M~T+ strain on M + T~ cells 
— does not produce recombinants. The 
two strains differ, therefore, in donor ca- 
pacity. 

The M + T~ donor strain (but not the 
M~T + strain) is found to harbor a phage. 
This virus, P22, is said to be nonvirulent 
or temperate. Nevertheless, about one in 
a thousand times this phage replicates and 
lyses or bursts the host cell, liberating up 
to several hundred progeny phage. Ac- 
cordingly, a culture of bacteria harboring 
temperate phage does not show a conspicu- 
ous amount of lysis. Because each cell of 
the M + T~ strain carrying P22 is poten- 
tially subject to lysis, the strain is said to 
be lysogenic. (The lysogenic bacterium, or 
lysogen, is immune to new infection — that 
is, to superinfection — by identical or ho- 
mologous phage.) On the other hand, the 
M~T + strain normally lacks P22 and is a 
nonlysogenic or sensitive strain. When a 
sensitive strain is exposed to temperate 
phage, a relatively large fraction of the 
newly-infected cells lyse and liberate phage. 
But a small fraction is able to survive, be- 
come lysogenic, and give rise to lysogenic 
progeny. If lysogens are lysed artificially 
and tested for phage, none are detected. 
Apparently, the phage in a lysogen is con- 
verted to a new form, called prophage, 
which reproduces at the same rate as the 
host chromosome. Usually, to lyse a lyso- 
gen, prophage must first rapidly replicate a 
number of times to produce the infective 
phage liberated at the time of lysis. 

What is the relationship between the fil- 
terable M+ factor and the phage P22? 

1. Both are unaffected by RNase and 
DNase. 

2. Both show the same inactivation pat- 
tern with temperature changes. 

3. Both have the same susceptibility to 
an antiserum that blocks the attachment of 
phage to the bacterium. 



4. Both become attached to susceptible 
cells simultaneously. 

5. Both have the same size and mass as 
determined by filtration and sedimentation 
tests. 

6. Both appear in the medium at the 
same time and in a constant ratio. 

7. Both retain this ratio even though var- 
ious purification and concentration proce- 
dures are applied. 

From these results, it is evident that M + 
is associated with the phage. Since the ge- 
netic material of Salmonella is known to be 
composed of DNA, it is likely that the ge- 
netic factor M+ is also composed of DNA. 
Moreover, because the M+ genetic factor 
cannot be located on the outer surface of 
the phage particle, the M+ gene must be 
located in the interior of the virus. 

Genetic transduction is the process of ge- 
netic recombination made possible by a vi- 
rus particle introducing homologous DNA 
into a recipient cell. 

Are there any restrictions on the genetic 
material of Salmonella which can be trans- 
duced by P22? This virus can be grown 
on sensitive bacteria genetically marked 
M + T + X+Y~Z~\ the crop of phage pro- 
duced after this infection can be harvested, 
and a portion tested on sensitive indicator 
strains (M~, T~ , X~, Y~, Z~) one at a 
time. The results of such tests show trans- 
duction of M + , of T+, and of X + — but not 
of Y + or Z + . Another portion of the har- 
vested phage is grown on another genetically- 
marked, sensitive strain — M + T~X+Y + Z-, 
for example. When the new phage crop is 
harvested and then tested on the indicator 
strains already mentioned, it is found now 
that the new crop of phage has lost T+ but 
has gained Y + transducing ability. These 
results demonstrate that a phage filtrate has 
a range of transduceable markers exactly 
equal to that of the markers present in the 
bacteria on which the phage was last grown. 



332 



( II \I»TER 25 



In other words, the phage is passive with 
respect to the content of genes it transduces 
and retains no transducing memoiy oi an\ 
hosts previous to the last. Since additional 
tests demonstrate that ever) locus in Sal- 
monella is transduceable by P22, we can 
call this a case of unrestricted or general- 
ized transduction. In generalized transduc- 
tion one cell is transduced for a given 
marker for about each 10''' infecting phage 
particles. 

Any chromosomal marker is transduce- 
able by P22. but is it possible to transduce 
more than one at a time? P22 can be 
grown on M + T+X + , harvested, and then 
grown on M~T~X-. The latter bacteria 
are replica-plated on three different media 

one selecting only for M+ recombinants 

(it contains T and X), another only for 
T+, and the third only for X l . When the 
M+ clones are further typed, they are still 
T-X~. Similarly. r+ clones are still 
M-X~, and X+ clones are still M~T~. 
These results show that only a single bac- 
terial marker or a relatively short DNA seg- 
ment is transduced at one time. In this re- 
spect, transduction is similar to transforma- 
tion but different from conjugation, in which 
— especially in Hfr strains — large sequences 
of genes can be transmitted and integrated. 
In Salmonella, however, examples are 
known of several genetic markers trans- 
duced together in what is called linked trans- 
duction or cotransduction. Other work has 
established that the biological synthesis of 
the amino acid, tryptophan, is part of a se- 
quence of genetically-determined reactions 
that proceed from anthranilic acid through 
indole to tryptophan. Different genes con- 
trolling different steps of this biosynthetic 
sequence are cotransduced ■; this finding 
suggests such genes are closely linked to 
each other. The biosynthesis of histidine in 
Salmonella is known to involve at least eight 
loci, four of which produce identifiable ef- 
• As shown by M. Demerec and coworkers. 



fects on the sequence of chemical reactions 
involved. Linked transductions have been 
found between two or more of these loci. 4 
In fact, using the relative frequencies of dif- 
ferent cotransductions and other evidence. 
all eight loci are found to be continuous with 
each other and to lie arranged linearly (see 
Figure 25-1). Using cotransduction, one 
can build up a complete and detailed ge- 
netic map of Salmonella which proves to be 
a single circle. Cotransduction of closely- 
linked markers is also known to occur " in 
E. coli by phage P 1 . 

In a generalized transduction experiment, 
when a prototroph is obtained by transduc- 
ing an auxotroph, the new prototroph is 
usually stable and produces clones pheno- 
typically identical to typical prototrophs. 
This process is called complete transduc- 
tion. In this case, the prototrophic gene 
introduced must have integrated into the 
Salmonella chromosome in place of the re- 
cipient's auxotrophic allele. However, in 
addition to the large prototrophic colonies 
formed on selective agar (each of these 
clones represents a complete transduction), 
on occasion about ten times as many minute 
colonies are present (see Figure 25-2). 
These minute colonies do not appear in 
platings of auxotrophic mutants on minimal 
medium and, also, are not the result of an 
interaction between auxotrophs and colonies 
of normal or transduced prototrophs located 
elsewhere on the plate. Minute colony for- 
mation is explained as follows: Through 
phage infection the cell initiating the minute 
colony receives the segment of DNA con- 
taining the gene for prototrophy under test. 
This gene, however, fails to be integrated 
and fails to replicate but retains its func- 
tional ability to produce a phenotypic effect. 
Consequently, a hybrid merogenote or het- 
erogenote is produced in which the domi- 
nant injected gene for prototrophy is func- 

4 By M. Demerec. P. E. Hartman. and coworkers. 
■ From the work of E. Lennox. 




1 

c 


' 1 1 


o 




CT> 


E 


■J 




(T 






o 


o 


E 




o 






c 


-C 


Op O 


O 


•— " 



a: 






»3 






333 






( ii \i> 1 1 u 25 




i K.i ri 25-2. Large and minute (period-sized) 
colonics of Salmonella, representing complete 
ami abortive transductions, respectively. (Cour- 
tesy of P. E. Hartman. ) 



tional. Because the prototrophic gene prod- 
uct is made, the cell is able to grow and 
divide. Only one of the first two daughter 
cells, however, receives the added chromo- 
somal fragment, the exogenote. The daugh- 
ter cell without the exogenote is able to grow 
and divide only until the prototrophic gene 
product received from the parent becomes 
too scarce; on the other hand, the hetero- 
genotic daughter cell can continue to grow 
and divide, in turn producing only one het- 
erogenotic daughter cell. In this way a mi- 
nute colony is produced which contains a 
single genetically-prototrophic cell. This has 
been proved in a variety of cases and by 
various methods.' This consequence of the 
failure of complete transduction is called 
abortive transduction. 

Hypothetically, the exogenote in an abor- 
tive transduction has two possible fates: the 



exogenote might eventually be lost; or it 
might be integrated, resulting in a complete 
transduction. Regardless of its ultimate 
fate, the exogenote is considered to be ge- 
netic in nature, even though it does not self- 
replicate. Remember, however, that sell- 
replication is an assumed characteristic of 
the total genetic material: this capacity was 
not required when a ^nc was first defined 
(p. 33). 

In most transduction studies, an excess 
of phage is used: that is, each cell is infected 
with more than one phage particle. In such 
experiments, it is always found that trans- 
duced cells become lysogenic simultaneously. 
Thus, the cell transduced receives not only 
the exogenote but an apparently-complete 
genome of a phage as well — the former re- 
sulting in genetic recombination for one or 
more host markers; the latter in lysogeny 
and immunity. The phage particle whose 
contents make the host cell lysogenic need 
not be the same particle which introduces 
the exogenote, since high concentrations or 
multiplicities of infecting phage are used. 
and, on the average, each host cell is pene- 
trated by the contents of two or more phage 
particles. Consequently, one particle might 
furnish the exogenote, and another might 
cause lysogeny and immunity. Using low 
concentrations of phage to obtain low mul- 
tiplicities so that almost no bacterium can 
be infected by more than one phage, it is 
possible to prove that, at least in some cases, 
only one phage particle is needed per trans- 
duction. It is found. 7 moreover, that when 
a single phage attacks a susceptible bac- 
terium, the virus can usually produce only 
one of three mutually-exclusive effects on 
its host — namely, lysis, lysogeny, or trans- 
duction. 

E. coli strain K12 is normally lysogenic 
for the temperate phage lambda (A). An- 
other strain of /:. coli is nonlysogenic, that 



,; By B. A. I). Stocker, J. Lederberg, and N. D. 
Zinder. and by H. Ozeki (1956). 



By J. N. Adams and S. E. Luria. 



Transduction 



335 



is, sensitive to lambda. This phage can be 
collected from a culture of lysogenic bac- 
teria in great quantities a few hours after a 
brief exposure of these cells to ultraviolet 
light. Such UV -induction causes the pro- 
phage to replicate, and the progeny phage 
lyse the cell. Using A phage collected from 
lysogenic cultures, one finds that only a very 
limited number of different bacterial mark- 
ers can be transduced. They are restricted 
to a region that controls galactose fermen- 
tation, the Gal locus, whose markers are 
known from conjugation studies to be very 
closely linked. Lambda is therefore capable 
only of restricted or specialized transduction. 

As mentioned, lysis of a lysogen and the 
consequent liberation of infective phage can 
be induced by ultraviolet light. When lyso- 
genic Hfr conjugate with sensitive F~, a 
number of zygotes are induced to lyse and 
liberate infective phage. Initiated by con- 
jugation this method of inducing prophage 
to produce infective phage and lysis, is 
called zygotic induction. Moreover, zygotic 
induction by A occurs, with a given Hfr 
strain, at a specific time after the start of 
mating. This precise timing suggests that 
the chromosome has a locus with which 
lambda prophage is physically associated 
during lysogeny. Jn a nonlysogenic cell, 
no prophage is attached to or associated 
with this site. When the site with the pro- 
phage enters a sensitive F~ cell, zygotic in- 
duction occurs. When crosses are made 
between nonlysogenic Hfr (without pro- 
phage) and lysogenic F^ (with prophage), 
however, zygotic induction does not occur, 
and the nonlysogenic locus is transferred and 
segregates in the heterogenotes just as any 
other genetic marker. From these results 
and others, the locus for lambda prophage 
maintenance is found to be closely linked to 
the Gal locus which lambda can transduce 
(see Figure 24-5, p. 324, in which the at- 
tachment locus is given as A). 

The original lambda-containing lysogenic 



K12 bacterium is stable and haploid with 
aspect to the Gal locus and produces only 
about one <7<//-carrying lambda per 10 (! -10 7 
phage. A lysate made by inducing such a 
lysogens is capable of producing only a low 
frequency of transduction ( LFT ) . Some of 
the cells transduced by an LFT lysate form 
clones unstable with respect to Gal, that is. 
that segregate out cells with the Gal geno- 
type of the recipient cell. In other words 
a Gal' bacterium, transduced by an LFT 
lysate of lambda carrying Gal+ , is usually 
an unstable heterogenote being diploid and 
heterozygous for Gal and occasionally seg- 
regating Gal- progeny. The merozygote 
produced by lambda transduction differs 
from that produced by P22 in an abortive 
transduction. In the latter case, the trans- 
duced segment cannot replicate; in the for- 
mer case, the transduced segment can rep- 
licate, so that clones of merozygotes can 
be produced. When infective lambda is in- 
duced from a lysogenic host merozygotic for 
Gal, the lysate contains one hundred times 
as many phage which carry a Gal locus as 
does the lysate of haploids. Such a Gal- 
rich crop of phage is capable of a high fre- 
quency of transduction (HFT). 

In the case of phages capable of gener- 
alized transduction, transducing phage can 
be obtained from the lysate of sensitive cells 
infected with free phage; for phage capable 
only of restricted transduction, transducing 
phage cannot be obtained this way. There- 
fore, in lambda, transducing phage are not 
found in the lysate of infected nonlysogen- 
ics and are released only from lysogenic 
(haploid or merozygotic) bacteria." 

By employing different multiplicities and 
combinations of transducing lambda (col- 
lected from lysogenic cells) and nontrans- 
ducing lambda (collected soon after nonly- 
sogenic cells are infected), it is possible to 

K The preceding account is based largely upon the 
work of J. Lederberg. E. M. Lederberg, and Nf. 1 . 
Morse, and of E. L. Wollman and F. Jacob. 



336 



CHAPTER 25 



prove ' that transducing lambda is defective 
for a portion of the lambda genome. The 
(idl locus being transduced replaces a \ari- 
ably-sized segmenl o( the lambda genome. 
The defective GaZ-transducing lambda par- 
ticle. A(/,e. retains certain phage properties 
and loses others. \dg has lost the abilit\ 
to replicate and produce infective phage 
progeny and the prophage it forms must be 
defective; the host cell infected with a sin- 
gle Adg particle is never lysogeni/ed. A 
cell infected by Adg, therefore, can be su- 
perinfected with a nontransducing phage 
whose additional presence makes the host 
lysogenic and contributes a function which 
permits the defective prophage to multiply 
after induction. At the time of lysis of such 
a doubly-infected cell, infective phages of 
both nontransducing and transducing types 
are liberated. This situation parallels that 
already described for Salmonella which can- 
not be lysed or lysogenized if infected by a 
single transducing P22 phage, but which 
shows one of these characteristics if the host 
is also infected with one or more normal, 
nontransducing P22 phage particles. 

Transformation does not ordinarily occur 
in E. coli, probably because of difficulty in 
DNA penetration. If the DNA of Adg were 
isolated and somehow introduced into the 
bacterium, however, one would expect this 
DNA to behave as a transforming principle 
with respect to the Gal locus. Even if DNA 
does not penetrate E. coli by itself, it might 
be capable of entering with infecting whole 
phage. Indeed, this does occur ,n ; that is. 
using nontransducing lambda as a carrier 
or "helper." DNA isolated from Adg is ca- 
pable of Gal transformation. 

From the discussion of nontransducing 
P22 and lambda, it should be clear that 
such temperate phages have two alternative 

'■> See W. Arber. G. Kellenberger. and J. Weigle 

(1957). and A. Campbell (1964). 

10 See A. D. Kaiser and D. S. Hogness (1960). 



pathways of action open to them upon in- 
fecting a sensitive bacterium: the phage 
either lyses or lysogenizes its bacterial host. 
As clearlj shown in the case of lambda, the 
infecting phage either remains in the cyto- 
plasm where it replicates taster than the 
chromosome and eventually lyses and liber- 
ates progeny phage, or it integrates in the 
chromosome where it resides as prophage 
and is replicated as a regular chromosomal 
marker. Accordingly, lambda and most 
other temperate phages are episomes. 

What is the basis for the difference be- 
tween the temperate phages capable of gen- 
eralized and those capable of restricted 
transduction? A restrictive-transducing 
phage usually has a specific chromosomal 
locus for attachment to the host chromo- 
some, a generalized-transducing phage has 
not. Assuming — correctly — that the phage 
genome is nucleic acid, it can be suggested 
that the nucleotide sequence held in com- 
mon between prophage and chromosome is 
shorter for the generalized-transducing 
phage than it is for the specialized trans- 
ducer. In this connection it is noteworthy 
that evidence has been obtained 1! that a 
portion of the lambda genome is homolo- 
gous to the E. coli chromosome, as revealed 
by the ability of their denatured DNAs to 
base pair with each other. Several experi- 
ments suggest that a prophage makes the 
host cell immune to further infection by 
homologous phage, by preventing not the 
penetration of the DNA but its replication. 
This action parallels the suppression of free- 
F replication by integrated F. 

Transduction by temperate phages has 
been found to occur also in Pseudomonas. 
Vibrio. Staphylococcus, and Proteus, and it 
would not be surprising to find transduc- 
tion occurring in a wide variety of other 
types of cells, including human. 

11 By D. B. Cowie and B. J. McCarthy, and by 
\1. H. Green. 



Transduction 337 

SUMMARY AND CONCLUSIONS 

Genetic recombination of loci of the bacterial chromosome can be mediated by tem- 
perate bacteriophages in the process of genetic transduction. 

The transduced segment can be derived from any region of the bacterial chromosome 
(as in generalized or unrestricted transduction) or from a narrowly limited region (as 
in specialized or restricted transduction). The DNA segment transduced may, by 
integration, replace a chromosomal marker of the host (as in complete transduction), 
or it may produce a merozygote, in which case the exogenote can still function, whether 
it can replicate (as Gal exogenotes in E. coli) or not (as the exogenote in abortive 
transduction in Salmonella). 

A transducing <£ lambda is defective in its own genome. The deficient portion is 
replaced by a small segment of bacterial DNA acquired at the time the prophage was 
induced in its last host. 

Most temperate phages are episomes which, when attached to the chromosome, have 
some characteristics resembling those of integrated F. 




Norton D. Zinder. about 1954. 



REFERENCES 

Arber, W., Kellenberger, G., and Weigle, J., "The Defectiveness of Lambda-Transduc- 
ing Phage" (in French), Schweiz. Zeitschr. Allgemeine Path, und Bact., 20:659- 
665, 1957. Translated and reprinted in Papers on Bacterial Genetics, Adelberg, 
E. A. (Ed.), Boston: Little, Brown, 1960, pp. 224-229. 

Campbell, A., "Transduction," pp. 49-89, in The Bacteria, Vol. 5, Heredity, Gunsalus, 
I. C., and Stanier, R. Y. (Eds.), New York: Academic Press, 1964. 

Jacob, F., and Wollman, E. L., "Spontaneous Induction of the Development of Bac- 
teriophage A in Genetic Recombination of Escherichia Coli K 12" (in French). 
C. R. Acad. Sci. (Paris), 239:317-319, 1954. Translated and reprinted in Papers 
on Bacterial Viruses, Stent, G. S. (Ed.), Boston: Little, Brown, 1960, pp. 336-338. 

Jacob, F., and Wollman, E. L., "Genetic Aspects of Lysogeny," pp. 468-500, in A 
Symposium on the Chemical Basis of Heredity, McElroy, W. D., and Glass, B. 
(Eds.), Baltimore: The Johns Hopkins Press, 1957. 



338 < ii\ri ii< 25 

Kaiser, A. D., and Hogness, I) S., "The rransformation of Escherichia Coli with 
Deoxyribonucleic \cid Isolated from Bacteriophage \dg," J. Mol. Biol., 2:392- 
415. 1960. 

Morse, M. I . 1 ederberg, I M., and I ederberg, J., "Transduction in Escherichia Coli 
k-12." Genetics, 41:142-156, 1956. Reprinted in Papers on Bacterial Genetics, 
Adelberg, E. \. il d.), Huston: Little, Brown. I960, pp. 209-223. 

Ozeki. H.. "Abortive fransduction in Purine-Requiring Mutants ol Salmonella Ty- 
phiniurium." Carnegie Inst. Wash. Publ. 612, Genetic Studies with Bacteria, 97- 
106, 1956. Reprinted in Papers on Bacterial Genetics, Adelberg. E. A. (Ed.), 
Boston: Little, Brown, I960, pp. 230-238. 

Stent. G., Molecular Biology of Bacterial Viruses, San Francisco: Freeman & Co.. 1963. 

Wollman, F. F.. and Jacob. F., "Lysogeny and Genetic Recombination in Escherichia 
Coli K 12" (in French). C. R. Acad. Sci. (Paris). 239:455-456, 1954. Translated 
and reprinted in Papers on Bacterial Viruses, Stent. G. S. (Ed.). Boston: Little. 
Brown. I960, pp. 334 335. 

Zinder. N. D.. "'Transduction' in Bacteria." Scient. Amer.. 199:38-43. 1958. 

Zinder. N. D.. and Lederberg, J., "Genetic Exchange in Salmonella," J. Bact., 64:679- 
699. 1952. 

QUESTIONS FOR DISCUSSION 

25.1. How would you deline the term provirus? How do the terms merozygote and 
heterogenote differ? How would you define a homo°enote? 

25.2. What characteristics are conferred upon a host cell infected by a nontransducing 
temperate phage which becomes a prophage? Does not become a prophage? 

25.3. How would \ou prove that only one exogenote exists in a microcolony of 
Salmonella produced by an abortive transduction? 

25.4. Discuss the statement: "Temperate phage has chromosomal memory, and the 
chromosome has temperate phage memory." 

25.5. F particles are known which carry the prophage of A as "memory." How could 
you prove the existence of such a particle? 

25.6. Describe the procedure and genotypes you would use in demonstrating that 
E. coli can undergo genetic transformation with respect to Gal. 

25.7. Is there any reason to believe that the close linkage of genes with related effects 
might be more advantageous in microorganisms than in higher organisms? 
Explain. 

25.8. List the different ways that the Gal locus in the E. coli chromosome can undergo 
recombination. 

25.9. Are temperate phages good or bad for bacteria? Explain. 

25.10. Is a cell which has presumably stopped undergoing mutation, genetic recom- 
bination, and self-replication of its DNA still considered to contain genetic 
material? Explain. 

25.11. Discuss the origin and relative numbers oi Adg present among the phages in 
l.FT and HFT lysates. 



Chapter 26 

BACTERIOPHAGE: RECOMBINATION 
AND GENETIC MAPS 



Tl 



I he morphology of the T-even 
group (T2, T4, T6) of phages 
that attack E. coli has been 
studied in some detail. 1 Its members are 
tadpole shaped, 0.1 to 0.2 p. long — roughly 
a tenth the bacterial diameter (Figure 26- 
1 ) . The surface of the head has a hexa- 
gonal outline and facets like a crystal. The 
head membrane is composed of numerous 
subunits each having a molecular weight of 
about 80,000. The tail is cylindrical and 
is used by the phage for attachment to the 
host cell. The outer sheath of the tail, com- 
posed of about 200 spirally-arranged sub- 
units each having a molecular weight of 
approximately 50,000, forms a hollow cylin- 
der. The sheath can contract, shortening 
its length while increasing its diameter with- 
out changing its volume appreciably. Be- 
neath the sheath is the core, a hollow cyl- 
inder with a central hole about 25A in 
diameter. At the distal end of the core is a 
hexagonal plate to which six tail fibers are 
attached; each is bent in the middle and 
seems to contain subunits with molecular 
weight of not less than 100,000 a piece. 
The subunits of the head membrane, sheath, 
and tail fibers are composed of protein. 
When digested with trypsin, each of these 
subunits produces a unique set of peptides 
indicating that each is different. The core 
is also protein. A serologically distinct pro- 
tein, 4 to 6% of the total phage protein, 
is found in the interior of the phage par- 

1 See S. Brenner et al. (1959). 
339 



tide; polyamines, putrescine, spermadine, 
lysozymc. and a minor polypeptide are also 
reported in the phage interior. 

In addition to these components, the T- 
even phage interior contains DNA whose 
volume is approximately the same as that 
of the total protein. This DNA is com- 
posed of a single double helix about 200,- 
000 nucleotides long. Since such a poly- 
nucleotide would be about 68 p. long, the 
DNA inside the phage must be highly 
coiled.- No RNA has been reported in 
DNA-containing phages. 

Not all phages contain DNA; several bac- 
teriophage contain RNA and no DNA. 
Moreover, the physical and chemical com- 
plexity of the T-even phages is not typical 

-See R. Kilkson and M. F. Maestre (1962). 



700 A 



HEAD 



MAIL 




figure 26-1. Diagrammatic representation of 
the structures observed in intact and triggered 
T-even phages of E. coli. 



340 



CHAPTER 26 



oi all viruses. Certain plant viruses, such 
is the tobacco mosaic and turnip yellow 
mosaic, arc relative!) simple helical or 
spherical structures. Although </>X174 
seems to have a simple spherical structure. 
4>R, a closely related single-stranded DNA 
phage, shows a small knob which may func- 
tion as a tail. 

Identification of the genetic material in 
DNA-containing phages is made somewhat 
easier because DNA contains no sulphur 
and T2 phage protein contains no phospho- 
rous. The DNA in one sample of phage 
can be labeled by feeding the E. coli host 
cells radioactive P : -, while the protein in 
another phage sample is labeled by feeding 
the host cells radioactive S 35 . Each sample 
of radioactive phage is then permitted to 
infect nonlabeled cells. '■ The following re- 
sults are obtained: in one sample all of the 
P 32 (hence all of the DNA) enters the bac- 
terium; in the other all but about 3% of 
the S 35 (hence almost all the protein) re- 
mains outside and is removed by blendor 
treatment. As implied earlier (p. 330), 
when most of the protein of an attached 
phage is removed from the host cell by 
blendor treatment the normal outcome of 
infection remains unaffected. These results 
are consistent with the view that DNA and 
not protein is the carrier of phage genetic 
information. 

; This account follows the work of A. D. Hershey 
and M. Chase ( 1952). 



Recall (p. 336) that pure DNA does not 
penetrate normal E. coli unassisted. The 
cell wall of /.. coli can be removed by suit- 
able culture conditions, leaving a protoplast 
that can be penetrated by purified DNA. 
After phenol, CaCT, or other treatments, 
the entire protein coat of phage can be re- 
moved, leaving pure DNA. When proto- 
plasts of /•-'. coli are mixed with such pure, 
single-stranded DNA of phage XI 74 (which 
has only about 4,500 dcoxyribotides per par- 
ticle), typical <£X174 progeny, including the 
characteristic protein envelope arc pro- 
duced. 1 Consequently, the only genetic ma- 
terial in DNA-containing phage is DNA. 

The course of events leading to lysis of 
a phage-infected bacterium can be sum- 
marized as follows (Figure 26-2): The 
phage becomes attached tail-first to specific 
receptors on the bacterial surface. All the 
DNA and a small amount of protein are 
injected into the host; probably, the injec- 
tion is assisted by the contraction of the 
spiral sheath protein. An eclipse period 
follows (Figure 26-2, B-D), during which 
no infective phage can be recovered if the 
host cell is artificially lysed. During this 
eclipse period, the infected cell is said to 
carry immature phage, and the phage DNA 
is replicating to produce a pool of phage 
DNA units. Starting at the end of the 
eclipse period (Figure 26-2E), a fraction 

1 This has been shown by G. D. Guthrie and R. L. 
Sinsheimer. 



FIGURE 26-2 (opposite) . Electron micrographs of growth of T2 virus inside the E. coli 
host cell. A. Bacillus before infection. B. Four minutes after infection. C. Ten min- 
utes after infection. The thin section photographed includes the protein coat of T2 
which can he seen attached to the bacterial surface. D. Twelve minutes after infection. 
New virus particles are starting to condense. E. Thirty minutes after infection. More 
than 50 T2 particles are completely formed and the host is about ready to lyse. (Cour- 
tesy of E. Kellenberger. Reprinted from the Scientific American, 204:100, 196E) 



Bacteriophage: Recombination and Genetic Maps 



341 









:ui 



CHAPTER 26 



of this DNA pool is assembled into mature 
phage, each genome surrounded by a newly 
synthesized coat (head and tail) (Figure 

26 2D). About 20 to 40 minutes after in- 
fection, the infected bacteria produce en- 
zymes called endolysins which rupture the 
bacterial cell wall and liberate infective 
phage into the medium. 

This last step completes the lytic cycle 
of a bacteriophage — the only one possible 
for intemperate or virulent phages, such as 
T. For temperate phages, this is one of 
the two possible cycles — upon entering a 
bacterium, the alternative is integrating with 
the bacterial chromosome as a prophage, 
thereby making the bacterium lysogenic and 
immune. Even in this event, remember, 
prophage occasionally dissociates from the 
chromosome and replicates to produce in- 
fective phage liberated by lysis of the host 
cell. 

Virulent Phages 

Methods for assaying the amount of phage 
present in a solution are based upon the 
virus's capacity to lyse sensitive bacteria. 
In one commonly-used method, the surface 
of an agar-containing plate is heavily seeded 
with sensitive bacteria which, upon incuba- 
tion, will grow to form a continuous and 
somewhat opaque lawn. When a few in- 
temperate phage particles are mixed with 
the sensitive bacteria before incubation, 
each particle enters a different bacterium, 
grows there, subsequently lyses the host, 
and releases up to several hundred daugh- 
ter phage. These particles proceed to at- 
tack bacteria near the original host, caus- 
ing them to lyse later. The repetition of 
this cycle produces a progressively increas- 
ing zone of lysis that is detected as a clear- 
ing or plaque in the bacterial lawn. Under 
these conditions each plaque is derived from 
one ancestral phage and a count of plaques 
therefore corresponds to a count of phage 
in the infecting sample. 



The detailed appearance of a plaque de- 
pends upon the medium, host, and phage. 
When other factors arc controlled, dilTcrent 
mutants of a phage may produce plaques 
with characteristically dilTcrent morphology. 
Plaque differences can involve size, turbid- 
ity, presence or absence of a halo, nature 
of the edges, and — when a dye is added to 
the agar in the plate — color. The investi- 
gator can. therefore, detect and maintain 
phage mutants affecting plaque type. Ge- 




figure 26-3. Plaques produced by parental 
and recombinant phage types. Progeny phage 
of a cross between hr+ and h ' r were tested 
on a mixture of suitable indicator bacteria. 
The small clear and the large turbid plaques 
are made by the parental types of phage prog- 
eny (h r f and h • r, respectively). The large 
clear and the small turbid plaques are produced 
by the recombinant types of progeny (h r and 
h + r+, respectively) . (Courtesy of A. D. 
Hershey.) 



Bacteriophage: Recombination and Genetic Maps 



343 



netically, phages can differ according to the 
hosts they are able to infect, and mutants 
occur which change the range of hosts at- 
tacked. Therefore, phage mutants affect- 
ing host range can also be detected and 
maintained. 

One can obtain a strain of intemperate T 
phage that is mutant both for host range, 
h, and plaque type, r. Sensitive bacteria 
are infected by a single phage particle con- 
taining both markers to determine the muta- 
tion frequencies to the wild-type alleles (h + 



and r+); wild-type phages are used to de- 
termine the mutation rates to each of the 
two kinds of mutant alleles. When the sen- 
sitive bacterial strain is exposed to a highly 
concentrated mixture of the double mutant 
(h r) and wild-type (h + r+) phages, so 
that some of the multiply-infected cells carry 
both phage types, not only do the parental 
types (h r and h + r+ ) occur among the 
progeny, but the recombinant types {h r 
and h r+ ) occur in frequency too high to 
result from mutation (Figure 26-3). Con- 




figure 26-4. A recombination map of T4D. Filled-in areas represent min- 
imal lengths for genes. The symbols for phage components represent the 
typical morphological products present in lysates of mutant-infected E. coli. 
(Courtesy of R. S. Edgar; see F. W. Stahl, et al.. Genetics, 50:539-552, 1964.) 



:M4 



chapter 26 



sequently, such experiments prove thai ge- 
netic recombination occurs between phage 
panicles in a multiply-injected cell. From 
the relative frequencies with which differenl 
recombinants appear among phage released 
Erom cells multiply-infected with a series of 
different mutants (this procedure is known 
as "crossing" genetically-different phages), 
the genetic map of a phage can be con- 
structed. When this is done for T4. the 
mutant loci are found to be arranged in a 
single closed linear order, that is. a circle 
(Figure 26-4). 

Because the T4 plaque mutant r lyses 
rapidly, it produces a larger plaque with 
sharper margins than the phage with the 
wild-type allele, r Mixed infections with 
r and r+ phages usually yield progeny that 
produce plaques of one or the other type. 
Two per cent of the observed plaques, how- 
ever, are mottled; that is, they appear partly 
r and partly r + . When mottled plaques 
are picked and their phage content tested, 
they produce progeny that make either r or 
r" type of plaque. Since both parental 
types are present, the mottled plaque could 
not have been initiated from a single phage 
haploid for the r locus. Mottled plaques 
are not caused by infection with clumps of 
phage particles; moreover, these plaques are 
not initiated by a phage carrying an un- 
stable r mutant, since unstable r phages pro- 
duce phenotypically — and genotypically — 
sectored, not mottled, plaques. From these 
results and others, it has been proved 6 that 
the two per cent of T4 phage producing 
mottled plaques in mixed infections are het- 
erozygous for a short region of the phage 
genome that includes the r locus. 

A single phage particle can be hetero- 
zygous for several loci, provided they are 



GENOTYPE 

rl or rill Mutants 

rll Mutants 
-+ 



PLAQUES FORMED 
ON HOST STRAIN 

B K 

r r 

r None 

+ + 



i k.i RE 26-5. Behavior of r mutants of T-even 
phages in the B and K strains of E. coli. 



located far enough apart. In fact, it is un- 
likely that any phage is completely haploid. 
Regions present in diploid condition are 
said to be redundant. Redundancy appears 
to be accomplished in two ways 7 : Either 
both regions are part of one DNA double- 
helix {terminal redundancy) or the extra re- 
gion is present as a separate segment of 
double-helix DNA {internal redundancy) . 

Genetic Fine Structure of <£T4 B 

The r mutants occur in three distinct re- 
gions of the T4 genetic map — rl, rll, and 
rill. The r mutants in all three regions 
produce plaques when E. coli strain B is 
used as host. However, mutants in the rll 
region are unique in that they cannot form 
plaques when their host is strain K.12 of 
E. coli (which happens to be lysogenic for 
lambda), whereas the rl and /•/// mutants 
and r+ phages can (Figure 26-5). Thus, 
among r mutants, only those in region II 
have this restriction in host range. 

The host-range restricted rll mutant is 
useful since it can be employed as a selec- 
tive marker. The mutation frequency from 



After the work of M. Delbruck and W. T. Bai- 
ley, and of A. D. Hershey and R. Rotman. 
'•See A. D. Hershey and M. Chase (1951); see 
also A. H. Doermann and L. Boehner (1963). 



7 See G. Streisinger, R. S. Edgar, and G. H. Den- 
hardt (1964). 

s The following discussion is based mainly upon 
the work of S. Benzer (1955, 1957). 



Bacteriophage: Recombination and Genetic Maps 



345 



/•// to r//+ can be determined readily by 
plating T4/7/ on strain K12, since only mu- 
tants to /•+ will form plaques (r+ is "se- 
lected" on strain K12). A large number of 
rll mutants that have a low mutation fre- 
quency (sometimes as low as one per 10 s 
phages) can be obtained. T4/7/ mutants can 
be divided into two classes, A and B, on the 
basis of their behavior after mixed infection 
of strain K12. When K12 is infected with 
an rll phage from each class, growth of the 
phage and lysis of the host occurs. This 
behavior suggests that the rll region is com- 
posed of two subregions, A and B, and the 
products of both are required to produce the 
normal r+ phenotype. Mutants defective 
only in the A subregion presumably can still 
make normal B product, and vice versa. In 
a bacterium multiply-infected with one phage 
mutant in A and another in B, the B and A 
products produced by the mutants can co- 
operate — that is, show complementation — to 
produce the r+ phenotype (Figure 26-6). 
If the two different rll mutants in a mul- 



tiple infection of strain K12 are located in 
the same subregion — region A, for example 
— they will be unable to produce the r + 
phenotype by complementation since neither 
phage can produce normal A product. In 
such cases the phage cannot grow and the 
host will not lyse unless the infecting mu- 
tants have lesions far enough apart in region 
A to permit wild-type progeny to result from 
recombination between them. 

Two mutants, rl and r2, arising independ- 
ently in the same subregion. may fail to re- 
combine with each other; however, rl may 
recombine with a third independent mutant, 
r3, even though mutant r2 does not. These 
results suggest that mutant r2 has a long de- 
ficiency or deletion that includes all or part 
of the region defective in rl and r3. Such 
deletion mutants are never found to revert to 
r+. Other mutants which do revert and 
which give no evidence of having long de- 
ficiencies are considered to be point mutants. 
Of the more than 1500 spontaneously-occur- 
ring rll mutants which have been typed, 



FUNCTIONAL COMPLEMENTATION 

A B 

x 



x y 

r X r 



» 



NO COMPLEMENTATION 



figure 26—6. The occurrence 
or nonoccurrence of comple- 
mentation between different 
rll mutants. 



X z 

r X r 



346 



CHAPTER 26 



i k,i ki 26—7. Genetic map <>/ the ill region o) phage T4. 
The breaks in the map militate segments as defined by 
the ends ot deletions. The order oj the segments has been 

determined as shown. I he order ot mutants within any 
one segment has not been determined, but all give recom- 
bination with each other. The hollow circles and other 
tilled-in symbols represent different types of phenotypic 
effects. (Courtesy of S. Benzer and S. P. Champe, Proc. 
Mat. Acad. Sci., U.S.. 47:1030-1031. 1961.) 



A CISTRON O^H^OO 



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130 129 80 



vf 



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1008 70e G O 



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P 5b9 IV 147 1034 632 

33 11 



H NB UV NT F En 19H A 
91 67fc9 272 392 88 S 12 



3099 
1176 



**. 



^ooooo- ooojr hCKX>B- KKX) — o- 



8C 380 UV NT 



BC 117 AP HB 



577 113 UV 159 



-o- 

E0 



-o- 




F P 635VJ SN )<H>S G 1310 585 (lb C 
70 79 AP 181 lb 135 

13b 



-o-o -o -o- -o- 



NT 
470 



-ooooo -ooo -o- -ooo 



385 157 1181 583 P 
3fa 



IGbO 566 EM 
71 




Wl C 311 279 531 13)6 1702 G JO 59613] 1314 H8 37] I55 
I82, ISl 11 129 



.'85 NT J 
518 3JI 




-oooo- o- -o 



P NT UV 
98 508 231 



Bfc 



65 B4b2 



<K>^(XKKKKK)^CX>- 



P I960 NT 979 NB 
83 281 68; 



EDb U/ 
50 IH 



Bt 



lala 



Blalb 



Bacteriophage: Recombination and Genetic Maps 



347 



-¥o- -CK><>0~CK>0-0<>0<K><><>^K><X 

O AP P 80 607 P 4 |J/ 289 N 681 I27H NT 1249 331 1562 AP P EM 



KX>0<>CK><><>0 

NA N AP SN NT 1011 SI) 1814 H 

27 7H 3\% 47 341 23 ,221 EM 100 3 24 11 249 58 55 

14 



Ml 58 24 




-o-o 



IJO 



-O -<>0000-O0-0-OOCK> -OO-OO -( 

486 228 1430 NT A N F MP & EM H NB 658 N NT NT NT 1518 NT 

97 56 84 102 56 164 52 42 4777 14 71 313 108 88 



<*P.\ 



HB309 



-ooooo- -o o-ooo -oooooooA- -ooo-oaoo^,, 

P UV 511 J H 547 50 G 1883 %5 H 1513 425 804 603 1470 UV N F UV EM UV 1130 733 447 ^^ 



14 1 255 201 



18 178 



14 



79 II 16 214 20 94 



018 

lOAPbt 
101 




-o- -o -oooooooo -oAo -CKXT 



^C/u 



H F 795 960 KB EH U J 1945 1345 573 P 2215 103 1322 UV 1404 2232 AP EM H 865 SD 8 

235 31 1112 50 548 148 5 122 76 64 51 61 72 



-o-o -oooo 

50 C 5D EM N 110 

145 36 131 135 21 



B CISTRON 




1071 qn J44 UV 3b0 



010 
072 



-oo^oAoooooo -o-ooooooo -ooo oo 



-^^556 



117 N 263 NT B 858 uv AP SD P J 
17 150 44 63 176 6 71 110 

BHbl 



1870 P 8 G P 114 P 5N 261 8 188 370 441 

4 55 19 87 42 103 



BHo.2. 



BHal 



B3 



84 8 

O EM 6 



-Boo -o -o 

OAP AP J0I6 50 EM 

35 S3 13 8 7 



Blal B<?b BlOal 8l0aX BlOb 



348 



CHAPTER 26 



about 300 arc different; that is. each is sepa- 
rable from all the others b\ recombination 
(Figure 26-7). Using overlapping deficien- 
cies and point mutants, it is possible to ar- 
range all the mutant loci of the A and B 
subregions in a single linear sequence with 
distances between mutants being approxi- 
mately additive. Thus, even in its fine struc- 
ture (within one "gene") the genetic recom- 
bination map of bacteriophage is linear. 

The ability to select for r + reversions by 
plating rll mutants on strain K12 permits the 
detection of mutation frequencies as low as 
one in 10\ This method also has approxi- 
mately the same efficiency for detecting re- 
combinants. Although numerous mutants 
were crossed with each other, the smallest 
reproducible frequency of recombination 
found between two mutants (0.02%) was 
relatively large, being at least one hundred 
times greater than the lowest frequency de- 
tectable. For recombination, therefore, 
0.02% seems to be close to the lower limit, 
a value which may be useful in estimating 
how finitely divisible DNA is for purposes of 
recombination. 

To calculate the smallest nucleotide dis- 
tance between two markers of T4 which are 
able to recombine with each other, we must 
assume that: 

1. The probability of genetic recombina- 
tion is constant per molecular distance on 
the phage genetic map. 

2. The genetic markers studied are repre- 
sentative of all other loci. 

3. The total length of the genetic map is 
accurately estimated by the summation of a 
number of small distances. 

Based on these assumptions, the total 
genetic map of phage T4 is calculated to 
be approximately 2500 recombination units 
long; that is, it shows 2500% recombination 
with respect to its total genetic content. 
(Remember that a recombination map based 
upon crossing over can be longer than one 



hundred recombination units — in this case, 
crossover units.) The molecular weight of 
the DNA of T4 is 130 to 160 times 10 ,; 
which means 14 contains about 400,000 nu- 
cleotides. These are arranged in the form 
of a single double helix of 200,000 linearly- 
arranged nucleotides. 

rr U c t . 2500% recombinants , 

The fraction A -^x ; —, — equals 

200,000 nucleotides 

0.0125% and expresses the percentage of 
recombination per nucleotide pair. Assum- 
ing that recombination cannot take place 
within a nucleotide pair, a phage genome has 
200,000 internucleotide points where ex- 
changes can occur. Thus, we can say that 
if two r mutants — different from wild-type in 
single, adjacent nucleotides — are crossed, r+ 
recombinants arc expected to occur among 
0.0125 % ^ Qr 0006 25%, f their progeny 

(the undetected double mutant recombinant 
also occurring with this frequency). 

Suppose that the lowest r+ recombinant 
frequency observed, 0.02%, is actually the 
minimal rate. This might mean that recom- 
bination could occur in each internucleotide 
position, but that the closest of the numerous 
mutants tested still would be separated by 
about three nucleotides ( .02 /.00625 ) . Sup- 
posing the mutants actually affect adjacent 
nucleotides, then only about every third 
internucleotide point, on the average, would 
be capable of undergoing recombination, and 
an estimate of the average unit of recombi- 
nation would be three nucleotides in length. 
Because the observed value of 0.02% is a 
maximum value for the least amount of re- 
combination, and uncertainties exist concern- 
ing the length of the genetic map and the 
number of nucleotides in the phage genome, 
the average nucleotide length between phage 
recombination events is subject to consider- 
able error. Since the DNA backbone does 
not seem to have any structural feature 
which occurs only every third nucleotide, it 
is reasonable to accept as a working hypoth- 



Bacteriophage: Recombination and Genetic Maps 



349 



esis that the smallest recombinational unit in 

phage equals one nucleotide.'" 

The entire rll region contains about 2000 
linearly-arranged nucleotides. Consider the 
functional characteristics of this region. If 
only the production of the r and r + pheno- 
types were considered, the rll region would 
behave as a single functional unit. The rll 
region, however, is composed of two sub- 
regions, A and B, which show complementa- 
tion. Such a division suggests that A and B 
are independent, separate units at the func- 
tional level. 

When E. coli strain K12 is doubly-infected 
with wild-type ( + + ) and T4 doubly-mutant 
(a x a- 2 ) in the A (or B) region, the r+ pheno- 
type is produced. In this case, the mutants 
are present in the same DNA double helix, 
being in the cis position (Figure 26-8). 
When the strain K12 bacterium is doubly- 
infected and each virus particle carries one 
of these mutants (a x -\-, +a 2 ), the mutants 
are in the trans position. No complementa- 
tion occurs, and no plaque is produced. 
When such a cis-trans test gives this result, 
the two mutants failing to complement in the 
trans position are said to belong to the 
same functional (A or B) unit, or cistron. 
The closest mutational sites between the A 
and B cistrons (Figure 26-6) are no more 
than 0.4 map units apart, indicating that the 
two cistrons are not separated by a large 
amount of DNA within which recombination 
can occur. 

Temperate Phages 

Genetic recombination also occurs between 
temperate phages. From the frequencies 
of genetically-recombinant phages occurring 
among the progeny of sensitive cells multiply 
infected with different mutants of lambda, 
it is possible to arrange the mutants on a 
single linkage map, just as is done for dif- 
ferent mutants in the virulent T phages. The 

9 Support for this view is found in work by D. R. 
Helinski and C. Yanofsky (1962). 



map for A, however, is (like T5) not circular 
as is the map for T4. 

The difference between a temperate and 
an intemperate phage is that the former can 
lysogcnize its host. When temperate phages 
infect sensitive bacteria, the plaques pro- 
duced have a turbid center caused by the 
growth of the bacteria that were lysogenized 
— not lysed. In such a temperate strain, 
mutants occur whose capacity for lysogeniza- 
tion is either decreased or lost (in which case 
the phage become virulent) and are detect- 
able because they form less turbid or clear 
plaques. Matings between phages that carry 
mutants with different degrees of virulence 
and other markers show that the loci con- 
trolling ability to lysogenize are a regular 
part of the phage genetic map. 

Some mutants seem to affect the very 
process by which phage is converted to pro- 
phage. Mutants of the latter type are called 
c {clear) mutants and occur in a cluster of 
loci within the phage genetic map. In 
lambda, three groups of c mutants occur in 
the sequence of c 3 , Ci, c- 2 (Figure 26-9). 
Mutants in the Ci segment no longer have 
any measurable ability to lysogenize and, 
therefore, cause completely-clear plaques, 
whereas those in c 3 and c 2 have about . 1 to 
.01 of the lysogenization ability of wild-type 
lambda. It has been demonstrated, further- 
more, that these three loci act at different 
times after infection. 

More than a dozen temperate phages in 



a, a 



1 "2 



+ + 



CIS 



+ a. 



TRANS 



figure 26-8. Cis and trans positions for two 
mutants in cistron rll A. 



.}.->() 



(HAITI K 26 



E. coli have been isolated. Some of these 

arc ultraviolet inducible, others are not. All 
(seven) of the viruses whose prophages arc 
inducible occupj different chromosomal loci 

located in one region of the host chronic 
some (Figure 26-10). These phages also 
differ from each other in that a host lysogenic 
for one can still be infected by any o\' the 
others 

Hemophilus influenzal' is transformable 
and can also be infected by the UV-inducible, 
temperate phage HP1. The genetic material 
o\ this phage can be introduced into nonlyso- 
genic Hemophilus competent for transforma- 
tion in three ways: by injection from intact 
phage; by infection with pure phage DNA; 
and by infection with pure bacterial DNA 

"' The preceding discussion is based primarily 
upon work of F. Jacob and E. L. Wollman 
(1957). of A. D. Kaiser and F. Jacob (1957). 
and of F. Jacob and A. Campbell. 



carrying the prophage, isolated from lyso- 
Lvns. In each case, the host cell either be- 
comes lysogenic or lyses and liberates mature 
phage progenj . 

Although the immunity properties oi' pro- 
phage are controlled by a small portion of 
the phage's total DNA content — the c, re- 
gion — this region alone docs not include all 
the genetic information or specifications 
needed for prophage to become mature, in- 
fective phage. Presumably most, if not all, 
the other phage loci arc essential for the pro- 
duction of mature phage, complete with its 
protein parts. 

Using unlabeled host cells, crosses can be 
made between genetically-different strains of 
lambda, one unlabeled and the other labeled 
with the isotopes C ,:: and N 15 . Following 
density-gradient ultracentrifugation, the dis- 

11 See W. Harm and C. S. Rupert (1963). 



d2i 
do 
di8 
di6 
di4 



di7 



d34 d22 C CO I 

d32 d30/' / / 




im 

FIGURE 26-9. Diagrammatic representation of the linkage group of the tem- 
perate bacteriophage X. The upper diagram shows the linear arrangement of 
various markers for host range or plaque size or type. The d symbols refer 
to specific defective mutants. The e region is marked by a thicker line and is 
shown enlarged in the loner diagram. It is composed of three sub-regions, 
(;,, (,, Co. Im refers to the segment controlling immunity. (After F. Jacob 
and J. Monod. ) 



Bacteriophage: Recombination and Genetic Maps 351 

tribution of labeled parental DNA is de- phages. The simplest explanation for this 

termined both among the parental and the result is that recombinants are formed after 

recombinant genotypes. In such experi- breakage and rejoining of the parental DNA 

ments, 1 - discrete amounts of the original strands. Such results do not support — but 

parental DNA are found in the recombinant do not exclude — phage recombinations oc- 
curring via a copy-choice mechanism accord- 



ing to which recombinant phage DNA is 
expected to be 
(1961). labeled material. 



12 See M. Meselson and J. J. Weisle (1961), and 

G. Kellenberger, M. L. Zichichi. and J. J. Weigle expected to be made of entirely new, un 



424 
Lac, Gal b 82 X 434 361 21 !466 

---I 1 I I I I I H- 

figure 26-10. Part of the E. coli linkage map showing the location of certain inducible 
prophages. 

SUMMARY AND CONCLUSIONS 

The morphology and lytic cycle of the virulent T-even phages of E. coli are discussed, 
and their genetic material identified chemically as DNA. After multiple infection 
with T4 phages carrying different genetic markers, genetic recombinants are found 
among the progeny. 

From the results of such phage crosses, a recombination map can be constructed 
for T4 in which the genes are arranged linearly in a single circle. Recombinant phages 
are often diploid for a short region between the recombinant markers. 

The genetic fine structure of the rll region of <£T4 is revealed by studies of mutation, 
complementation, and genetic recombination. Their data suggest the hypothesis that 
the smallest recombinational unit in phage is one nucleotide. 

Genetic recombination also occurs between mutants of a temperate phage. The 
single linkage group of lambda is not circular. Immunity to superinfection is deter- 
mined by the c x region of the phage genetic map. Sometimes, if not always, phage 
recombination involves parental strands which have broken and rejoined. 

REFERENCES 

Benzer, S., "Fine Structure of a Genetic Region in Bacteriophage," Proc. Nat. Acad. 
Sci., U.S., 41:344-354, 1955. Reprinted in Papers on Bacterial Viruses, Stent, 
G. S. (Ed.), Boston: Little, Brown, 1960, pp. 209-219. 

Benzer, S., "The Elementary Units of Heredity," pp. 70-93, in A Symposium on the 
Chemical Basis of Heredity, McElroy, W. D., and Glass, B. (Eds.), Baltimore: 
Johns Hopkins Press, 1957. 

Benzer, S., "On the Topography of the Genetic Fine Structure." Proc. Nat. Acad. Sci , 
U.S., 47:403-415, 1961. 

Benzer, S., "The Fine Structure of the Gene," Scient. Amer., 206 (No. l):70-84 
1962. 

Brenner, S., Streisinger, G., Home, R. W., Champe, S. P., Barnett, L., Benzer, S., and 
Rees, M. W., "Structural Components of Bacteriophage," J. Mol. Biol., 1:281- 
282, 1959. 



352 



CHAPTER 26 





A. D. Hershey. about I960. 



Seymour Benzer, in 1961. 



Davidson, P. F., Freifelder. D.. Hede, R., and Levinthal, C, "The Structural Unity of 
the DNA of T2 Bacteriophage," Proc. Nat. Acad. Sci., U.S., 47:1123-1129, 1961. 

Doermann, A. H., and Boehner, L., "An Experimental Analysis of Bacteriophage T4 
Heterozygotes, I.," Virology, 21:551-567, 1963. 

Guthrie, G. D., and Sinsheimer, R. L., "Observations on the Infection of Bacterial 
Protoplasts with the Deoxyribonucleic Acid of Bacteriophage ^>X174," Biochim. 
Biophys. Acta, 72:290-297, 1963. 

Hanafusa, H.. Hanafusa, T.. and Rubin, H., "The Defectiveness of Rous Sarcoma 
Virus, II. Specification of RSV Antigenicity by Helper Virus," Proc. Nat. Acad. 
Sci., U.S., 51:41-48, 1964. 

Harm. W., and Rupert, C. S., "Infection of Transformable Cells of Haemophilus in- 
fluenzae by Bacteriophage and Bacteriophage DNA," Zeit. f. Vererbungsl.. 94: 
336-348, 1963. 

Hayes, W., The Genetics of Bacteria and their Viruses, New York: J. Wiley & Sons, 
1964. 

Helinski. D. R., and Yanofsky, C, "Correspondence between Genetic Data and the 
Position of Amino Acid Alteration in a Protein," Proc. Nat. Acad. Sci., U.S., 
48:173-183, 1962. 

Hershey, A. D.. and ( hase, M., "Genetic Recombination and Heterozygosis in Bacterio- 
phage." (old Spring Harb. Sympos. Quant. Biol., 16:471-479, 1951. Reprinted in 
Papers on Bacterial Viruses, Stent, G. S. (Ed.), Boston: Little, Brown. 1960, pp. 
179-192. 



Bacteriophage: Recombination and Genetic Maps 353 

Hershey, A. D., and Chase, M., "Independent Functions of Viral Protein and Nucleic 
Acid in Growth of Bacteriophage," J. Gen. Physiol., 36:39-54, 1952. Reprinted in 
Papers on Bacterial Viruses, Stent, G. S. (Ed.), Boston: Little, Brown. I960, pp. 
87-104. 

Jacob, F., and Wollman, E. L., "Genetic Aspects of Lysogeny," pp. 468-500, in A 
Symposium on the Chemical Basis of Heredity, McElroy, W. D., and Glass, B. 
(Eds.), Baltimore: Johns Hopkins Press, 1957. 

Jacob, F., and Wollman, E. L., "Viruses and Genes," Scient. Amer., 204 (No. 6): 
92-107, 1961. 

Kaiser, A. D., and Jacob, F., "Recombination Between Related Temperate Bacterio- 
phages and the Genetic Control of Immunity and Prophage Localization," Virology, 
4:509-521, 1957. Reprinted in Papers on Bacterial Viruses, Stent, G. S. (Ed.), 
Boston: Little, Brown, 1960, pp. 353-365. 

Kellenberger, G., Zichichi, M. L., and Weigle, J. J., "Exchange of DNA in the Re- 
combination of Bacteriophage A," Proc. Nat. Acad. Sci., U.S., 47:869-878, 1961. 

Kilkson, R., and Maestre, M. F., "Structure of 72 Bacteriophage," Nature (Lond.), 
195:494-495, 1962. 

Meselson, M., and Weigle, J. J., "Chromosome Breakage Accompanying Genetic Re- 
combination in Bacteriophage," Proc. Nat. Acad. Sci., U.S., 47:857-868, 1961. 

Mosig, G., "Genetic Recombination in Bacteriophage T4 during Replication of DNA 
Fragments." Cold Spring Harb. Sympos. Quant. Biol., 28:35-42, 1964. 

Rubenstein, I., Thomas, C. A., Jr., and Hershey, A. D., "The Molecular Weights of T2 

Bacteriophage DNA and its First and Second Breakage Products," Proc. Nat. 

Acad. Sci., U.S., 47:1113-1122, 1961. 
Stahl, F. W., Edgar, R. S., and Steinberg, J., "The Linkage Map of Bacteriophage T4," 

Genetics, 50:539-552, 1964. 
Streisinger, G., Edgar, R. S., and Denhardt. G. H., "Chromosome Structure in Phage 

T4. I. Circularity of the Linkage Map," Proc. Nat. Acad. Sci., U.S.. 51:775-779, 

1964. 
See last portion of Supplement V. 



QUESTIONS FOR DISCUSSION 

26.1. Is the hole in the tail of T-even phages large enough for the passage of one 
or two double helices of DNA? Explain. In what respect does your answer 
concern the manner in which phage DNA enters into a bacterial host? 

26.2. In studying phages, what are the advantages of using bacteria growing on a 
solid, rather than in a liquid, culture medium? 

26.3. Do you think it would be feasible to study the genetic basis for different 
morphological or for different protein components of a phage? Explain. 

26.4. What is meant by a phage cross? Describe how you would know that you 
made one. 

26.5. Does the finding— that complete progeny phages are liberated after pure DNA 
from 4>X174 infects a protoplast — mean that all the information for making 
6X174 DNA and <f>X\14 protein is contained in the phage's DNA? Explain. 

26.6. Are the hypotheses of phage recombination by breakage and by copy-choice 
mutually exclusive? Explain. 



354 CHAPTER 26 

26.7. In what respects is d>\ similar to and different from an I particle? 

26.8. A temperate phage able to transduce any known chromosomal marker in E. coli 
is known. Would you be able to locate the chromosomal site for its prophage? 
I \plain. 

26.9. Does the finding — that a single phage particle can transduce a bacterial fragment 
carrying not onlj a bacterial marker hut two linked prophages — have any bear- 
ing upon the essentiality of the entire phage genome being present for infection 
and or the production o\ phage progeny? Explain. 

26.10. How can you distinguish a <£T4 mutant in the ill region from one in the rl or 
the rill region'.' 

26.1 I. Describe how the cis-trans test is used to show functional complementation be- 
tween two mutants in phage. 

26.12. What would you expect to be the near-maximum number of nucleotides trans- 
duceable by a phage still capable of phage activity? On what is your opinion 
based? 

26.13. If the average protein-specifying gene were 2000 nucleotides long, how many 
different proteins could be specified by ^>T4? By <£X174? 

2(^.14. What do you consider to be the most remarkable feature of </>X174? 

26.15. Mutants which show functional complementation in the pan-2 region of Neu- 
rospora can be arranged in the same linear order by complementation and by 
genetic recombination. Is it necessarily true that both maps will also be iden- 
tical for other regions? Explain. 

26.16. What is a functional genetic unit, or cistron? How is your answer related to 
its length in nucleotides? 

26.17. What have you learned in this chapter regarding the chemical scope of the 
genetic unit of recombination? Of function? 



Chapter 27 

BACTERIAL EPISOMES AND 
GENETIC RECOMBINATION 



T: 



(he location and genotype of 
F determines the kind of male 
sexuality which occurs in Esch- 
erichia. In this bacterium, conjugation leads 
to new combinations of either or both the 
chromosomal genes and the extrachromoso- 
mal episomal genes. 

Let us consider the sequence of events 
in F's genetic recombination. When an Hfr 
strain reverts to F + , the F particle integrated 
into the Hfr chromosome is somehow liber- 
ated from it, or deintegrated. The particle 
then enters the cytoplasm, replicates, and 
thereafter is infectious. In some subsequent 
generation, the F particle may reintegrate 
into a chromosome, making it Hfr. Before 
an F particle can integrate, it apparently must 
synapse with the chromosome — the attrac- 
tion between F and the chromosome prob- 
ably being the same as between a segment of 
donor chromosome and the host chromosome 
in transformation, conjugation, or transduc- 
tion. In transformation, the integrating do- 
nor loci must be homologous to those re- 
placed in the recipient cell. (Most likely, 
this homology is also required for the inte- 
gration of chromosome fragments introduced 
by conjugation or transduction.) Since F 
can integrate at a variety of loci, it appears 
likely that F has segments of DNA homolo- 
gous to a variety of chromosomal regions. 
The homologous segments which F presum- 
ably contains may have been present 'ini- 
355 



tially," or they may have been obtained at 
the time of previous deintegrations. (It is 
not known whether the first F originated as 
an offshoot of the bacterial chromosome or 
entered the bacterium from the outside.) If 
a free F particle sometimes carries an extra 
segment of chromosomal DNA somehow 
obtained at the time of deintegration, one 
should be able to find a type of free F parti- 
cle which, when introduced into an F - strain, 
shows a high affinity for a specific chromo- 
somal region. Recall that temperate phages 
capable only of restricted transduction show 
such a restriction, although the wild-type F 
particles do not. 

Consider next the following results from 
experiments l concerned with the expecta- 
tions mentioned. An F^ strain (carrying F 
extrachromosomally) gave rise to an Hfr 
strain, P4x, whose chromosomal markers 
were arranged in the following sequence: O 
(origin or lead point) -Pro-TL-Thi . . 
-Gal-Lac-SF (sex factor place of attach- 
ment). Crosses of P4x by F~ produced F~ 
progeny, except for Lac recombinants (which 
are usually Hfr males because of the close 
linkage of F and Lac). 

From P4x arose a new strain, P4x-1, hav- 
ing these characteristics: 

1. With respect to chromosomal loci, it 
was identical to P4x in the order of arrange- 
ment and in the times of entry (determined 
by interrupted conjugation). For example, 
both transferred Pro at about six minutes, 
TL at about twenty minutes, and Lac last. 

2. Recipients showed a frequency of re- 
combination for chromosomal loci lower 
than the frequency of recombination when 
P4x was the donor. For example. Pro re- 
combinants were 0.3 to 0.5% with P4x-1 
as donor and 4.8% with P4x as donor. 

3. Interrupted conjugations revealed that 

1 The following discussion is based primarily upon 
the work of E. A. Adelberg and S. N. Burns 
(I960), and F. Jacob and E. A. Adelberg (1959). 



356 



« ii \r iir 27 



mum of its recombinants for Pro 01 ll be 

haved us males. 

4. The male factor was linked neither to 
the above-mentioned loci nor to an) other 
chromosomal marker showing recombination. 

.v Like tree F, the male factor entered 
the F cell about five minutes after con- 
jugation began. 

6. Treatment with acridine orange elimi- 
nated the male sex factor, converting the 
cells to F - . 

These findings prove that the P4x-1 male sex 
factor — called F' — is located extrachromo- 
somally. 

Although F could attach at any one of a 
number of different chromosomal sites pro- 
ducing Hfr chromosomes differing in O point 
position and in direction of transfer. F' at- 
tached at a particular locus near Lac in such 
a way that chromosomal loci were always 
transferred in the same order and direction. 

Since P4x-1 transferred its chromosome 
more frequently than the typical F+ (F- 
containing) male, the chance of F' associat- 
ing with the chromosome near Lac seems to 
be greater than the total chance of F integrat- 
ing at any one of numerous different loci. 
P4x-1. on the other hand, transferred the 
chromosome less frequently than P4x, sug- 
gesting the possibility that F is not fully inte- 
grated in P4x-1 ordinarily, although it is in 
P4x. This difference would also explain why 
P4x-1 had free F\ whereas P4x did not. 
since chromosomal F prevented the estab- 
lishment of free F. Ordinarily, F' seems to 
exist as a kind of exogenote which synapses 
with the chromosome just before or after 
conjugation is initiated. The mechanism of 
chromosome mobilization seems to involve a 
recombinational event between the episomc 
F' and the chromosome. 

When F' was transferred to F~ cells as an 
extrachromosomal particle, the recipient cells 
were converted to males that, relativelv 



often, could transfer their chromosome in 
the same sequence as P4.\ and P4.\-l males, 
suggesting that the chromosome o[ the ordi- 
nary F cell has a segment of DNA near Lac 
homologous to a segment carried by F'; that 
is. F' possesses a chromosomal segment 
which can pair with a particular chromo- 
somal locus. 

As mentioned, treatment with an acridine 
dye eliminated the extrachromosomal F' par- 
ticles from the P4x-1 strain and converted it 
to F — . Such an F strain conjugates with 
males carrying either F or F' extrachromo- 
somally. In both cases the F strain was 
relatively often converted into a donor 
(male) that transferred its chromosome in 
the same sequence as P4x and P4x-1 males. 
Clearly, then, the F~ — derived from P4x-1 
via acridine orange — carries a chromosome 
which has retained a segment of F near Lac, 
the portion retained held in common by F 
and F'. In so far as the F portion of the 
particle is concerned, then, F and F' were 
not detectably different in these experiments. 
Since F' is found to be approximately twice 
the size of F, one can think of F' as being an 
F particle with an extra, particular piece of 
chromosome attached. 

The preceding suggests that F' can carry 
chromosomal DNA apparently still capable 
of replication in its new location. Let us 
suppose that this chromosomal segment is 
also still able to function normally. F' may 
fail to show a phenotypic effect for a normal 
chromosomal locus because it contains one 
or more (as yet unidentified) chromosomal 
markers. (Probably less than 1% of the 
chromosomal loci have been identified; more- 
over, there may be chromosomal regions 
whose only function is the maintenance of 
and recombination with episomes.) The 
very existence of F', however, encourages a 
search for still different F particles to which 
a known chromosomal marker might be 
attached. 



Bacterial Episomes and Genetic Recombination 



357 



Using Lac F cells and a Lac + strain of 
Hfr with F integrated very close to Lac 
rare recombinants are obtained from inter- 
rupted conjugations which receive Lac + too 
early. Certain of these recombinants have 
the following properties: 

1. They receive only F and Lac + . 

2. They are unstable and, occasionally, 
give rise to Lac~F~ individuals; hence, the 
original recombinant must be a merozygote 
carrying both Lac + and Lac~ alleles. 

3. When crossed to Lac~F~ cells, they 
simultaneously transfer both F and Lac + 
with 50% or higher frequency. This trans- 
fer starts soon after conjugation begins, just 
as in the case of free F (or F'), and is 
unlinked to other chromosomal markers. 
Thus, F-Lac+ behaves as a free single unit. 

4. The recombinant transfers its chromo- 
some in the same sequence as, but with a 
lower frequency (%o) than, the original Hfr 
line. These frequencies are exactly those 
found in comparing P4x-1 with P4x. 

5. The F-Lac + element can be trans- 
mitted in a series of successive conjugations, 
each recipient possessing the properties of 
the original recombinant. 

All these characteristics are most simply 
explained by an F particle which carries a 
chromosomal piece bearing Lac + deintegrat- 
ing in the original Hfr strain. The attached 
Lac + piece is known, moreover, to contain 
three cistrons governing the synthesis of /?- 
galactosidase, /3-galactoside permease, and 
the repressor for this system. From subse- 
quent integrations and deintegrations, one 
can also obtain F-Lac~ particles — composed 
of F-Lac with a Lac~ point mutation. 
Finally, another Hfr, with F integrated close 
to Pro, is found to produce an F-Pro particle 
whose properties are analogous to those of 
the F-Lac particle. If, however, an F-Lac + 
particle enters a cell containing a deletion in 



the Lac region, the particle, besides transfer- 
ring Lac ' , behaves like F in that it transfers 
random chromosomal markers with the low 
frequencies characteristic of ordinary F by 
F~ crosses. 

Since F can integrate at a variety of loci, 
these results suggest (and it turns out to be 
true) that upon deintegration any one of a 
variety of normal chromosomal loci can be- 
come a part of the genotype of cytoplasmic F 
and can replicate and function in the extra- 
chromosomal state. 

Particles like F' and F-Lac — substituted 
sex factors — represent a third type of male 
sex factor characterized by serving as inter- 
mediate donors of the usual chromosomal 
markers. It can be hypothesized that F, F' 
and other substituted sex factors are nor- 
mally small, double-helix, DNA ring chro- 
mosomes. If such small ring chromosomes 
were integrated in the E. coli chromosome 
by a single crossing over, the product would 
be a bigger (Hfr) ring. The now-integrated 
F particle would cause the enlarged ring 
chromosome to be open, usually at one end 
of F, and would mobilize the chromosome 
during conjugation. If the chromosome 
were opened at some internal position in F, 
part of F would be at the O point and part 
at the opposite terminus. Some evidence 
suggests that openings of this type sometimes 
occur. F may deintegrate by an internal 
crossing over, producing free ring F and the 
ring E. coli chromosome. It should be em- 
phasized that the suggested circular model of 
F is based upon no evidence and is nothing 
more than speculation at present. 

When F is not integrated, only one or a 
few F particles are present per E. coli chro- 
mosome — at least in cells that have carried 
F for some time. (This case is similar to the 
situation in which organisms with more than 
one chromosome are regulated by some 
mechanism in the cell which permits each 
chromosome to replicate only once a gen- 



;r>s 



CHAPTER 27 



eration.) Populations of donors, grown to 

saturation density in aerated broth or cul- 
tured on agar overnight, can lose their donor 
phenotype temporarily and behave as genetic 
recipients. Since they retain their sex factor 
yet behave as F cells, they are known as 
/•" phenocopies ( see p. 317). If a /^/c~F~ 
phenocopj carrying F is mated with an F^ 
male carrying F-Lac + , exconjugants can be 
obtained that carry both types of F. Soon, 
however, in some experiments, one or the 
other F particle persists in the progeny. This 
adjustment shows that there is some regula- 
tion of the number of F partieles per nuclear 
body. Hfr which are F~ phenocopies, do 
not tolerate the presence of an introduced 
autonomous sex factor. 

F not only mobilizes the entire chromo- 
some in Hfr cells, but it also mobilizes 
merogenotes — in an F-merogenote trans- 
fer - — as shown in the preceding discussion. 
This latter process has also been called sex- 
duction, F-duction, or F-mediated transduc- 
tion. With respect to such transduction, sub- 
stituted sex factors resemble Adg, just as F 
resembles A. One can construct a haploid 
Hfr containing two attached F factors so 
integrated that, upon mating, the chromo- 
some is present in two pieces — 1 4 and -'{. its 
length — each with an F at the end, and both 
merogenotes capable of being transferred 
to the F~ cell. An F-merogenote carrying 
the markers Pur, V6, and Lac transfers the 
merogenote so that the entry order — deter- 
mined by studies of spontaneous and artifi- 
cial interruptions of mating — is 0-Pur-V6- 
Lac-F. In all these respects, then, whole 
chromosome and merogenote mobilization 
by F appear identical, differing only with 
regard to the length of the genetic segment 
transferred. 

Although the merogenote markers can 

-' The remainder of this section is based upon A. J. 
Clark and E. A. Adelherg (1962). A. M. Camp- 
hell (1962). and W. Hayes (1964). 



sometimes integrate into the chromosome, 
F-linked merogenotes can also persist and 
replicate without integration, forming clones 
of merozygotes or partial diploids. The 
longer the merogenote — when it consists of 
49? or more of the genome — the more un- 
stable it is. 

Extrachromosomal F can pass from male 
to female during conjugation. Deintegration 
and integration of F can result in a two- 
directional flow of chromosomal genes be- 
tween F and the chromosome in the same 
cell. When F is at a chromosomal locus, 
the chromosomal genes are rendered mobile 
so that chromosomal genes can be trans- 
ferred to another cell. F can also enter an 
F~ cell by transduction, in which case it is 
recovered only in the free state, even if the 
donor was Hfr. (During phage growth, 
fragmentation of the chromosome takes place 
at or close to the ends of the integrated F 
element.) Consequently, the F particle is 
directly involved in genetic recombinations 
within and between bacteria which involve F 
itself and chromosomal genes. 

Promoters 

A promoter is a genetic element which pro- 
vides one or more of the special conditions 
needed for genetic transfer via conjugation. 
If a genetic element (like F) performs all the 
functions of a promoter, including mobiliza- 
tion of the entire chromosome or of a mero- 
genote, it is called a sex factor. Sometimes 
a promoter (such as F) promotes only its 
own transfer (into F~ cells). At other times 
a promoter and a whole chromosome or a 
merogenote are transferred in linkage. 

F-Lac promotes the transfer of both the 
merogenote and the chromosome. After 
ultraviolet treatment of F-Lac heterogenotes, 
some individuals are found no longer able to 
transfer Lac or chromosomal markers. Ap- 
parently a mutation in F resulted in a loss 
of one or more promoter functions. When 



Bacterial Episomes and Genetic Recombination 



359 



Salmonella or Shigella act as recipients in 
crosses with F+ or Hfr E. coli, F is trans- 
ferred but sometimes is unable to act as a sex 
factor until it is sent back into F~ E. coli. 
Such results show that promoter functions 
can be temporarily inhibited or unexpressed, 
depending upon the host genotype. 

* Resistance Transfer Factors 

A genetic agent — the resistance transfer fac- 
tor, RTF — has been found in Shigella. 3 
RTF is a promoter which causes conjugation 
and the transfer of a series of different, 
linked, drug-resistance genes; it is, therefore, 
a sex factor. Under special circumstances, 
RTF can mobilize chromosomal loci. It can 
be transferred from Shigella to Escherichia 
or Salmonella independently of the chromo- 
some, and such recipients can be cured of 
RTF by acridine dyes. RTF promotes its 
own transfer which starts within one minute 
of mixing the parents, and it replicates au- 
tonomously. Although RTF remains trans- 
ferable when introduced into an F + or Hfr 
cell, F-promoted chromosome transfer in Hfr 
cells is reduced one hundredfold, and F- 
promoted merogenote transfer and the trans- 
fer of free F are completely inhibited by cer- 
tain RTFs. When RTF is spontaneously 
lost, these F functions are restored. Other 
RTF strains have no effect on F function. 

F produces an antigen on the cell surface 
which must be present for <£f2 to attack 
males. When RTF and F are both present, 
a new RTF-antigen replaces the F-antigen. 
When a cell is infected by RTF its chromo- 
somal markers are mobilized one hundred 
times more frequently when its chromosome 
carries a segment of F than when it does not. 
These observations suggest that at least a 
partial genetic homology exists between RTF 
and F. In view of this and other evidence, 
it is concluded that RTF has some relation- 
ship with the chromosome, although it may 
3 See T. Watanabe (1963). 



not be able to assume a stably integrated 
state. RTF has been called an episome by 

its discoverers. 

* Colcinogenic Factors ' 

Many strains of enteric bacteria (Escher- 
ichia, Salmonella, Shigella, for example) pro- 
duce one or more highly-specific, antibiotic 
substances called colicins. Colicins are bac- 
tericidal but not bacteriolytic agents; a thou- 
sandth of a microgram of colicin can kill a 
million sensitive E. coli cells. More than a 
dozen groups of colicins are known; each 
group is designated by a different capital let- 
ter; each adsorbs to a different receptor site 
at the cell surface. Different colicins belong- 
ing to the same group can be distinguished 
by other characteristics. Colicins have a 
high molecular weight; two of them, colicin 
K and colicin V, have been purified and iden- 
tified as lipocarbohydrate-proteins. These 
purified colicins seem to be the same as the 
O antigen of the bacteria." 1 The antigen 
molecule can be separated into a lipocar- 
bohydrate and a protein fraction, the latter 
having all the colicidal activity. 

A cell able to produce a colicin is colicino- 
genic; one without this ability is noncolicino- 
genic. Colicinogeny is stable and can be trans- 
mitted through thousands of cell generations. 
Although it can be lost spontaneously, 
spontaneous acquisition of colicinogeny has 
never been observed. Consequently, it is 
reasonable to suppose that colicinogenic 
bacteria possess genetic material — colicino- 
genic (col) factors — that govern the syn- 
thesis of different colicins. Not only are 
these factors transmitted to progeny via vege- 
tative reproduction, but new strains can be- 
come colicinogenic through bacterial conju- 
gation or phage-mediated transduction. 

Only a small fraction of the bacteria in a 

4 This section follows the work of P. Fredericq 
(1963). See also W. Hayes (1964). 
"See W. F. Goebel (1962). 






CHAPTER 27 



colicinogenic culture actually produce coli- 
cins. Colicin is lethal to the bacterium syn 
thesizing it. but colicinogenic cells which do 
DOl \ icld colicin are viable and immune when 
exposed to corresponding colicin. Colicin 
synthesis can be induced in nearly all cells 
of a colicinogenic culture after exposure to 
ultraviolet light, nitrogen mustard, or hydro- 
gen peroxide. Thus, in several of the prop- 
erties mentioned, colicinogeny and lysogeny 
are similar, the col factors behaving in these 
properties like the prophages of temperate 
phage. When a strain is both lysogenic and 
colicinogenic, induction often releases either 
phage or colicin but not both. 

Bacteria can be protected against colicins 
via immunity or resistance. A colicinogenic 
bacterium, although immune to the corre- 
sponding colicin, can possess receptor sites 
which make it susceptible to colicins of other 
groups. Certain other genes confer resist- 
ance to whole groups of colicins by causing 
the loss of receptors. 

As noted, colicin K and the O antigen are 
apparently identical. When the receptor 
sites for colicins and for virulent phages are 
studied, in a number of cases colicin and 
phage are found to share receptor sites; for 
example, receptor sites are shared by colicin 
K and #T6; colicin E and ^BF-23; colicin 
C and <f>T\ or <^>T5. Since virulent phage 
attach to receptors by means of a protein 
located at the tip of their tails, colicin and 
tail-tip protein appear to he very similar. In 
serological tests, however, colicin and phage 
sharing a common receptor have not been 
found to exhibit any cross reaction, and 
colicinogeny does not confer immunity to 
infection by a site-sharing phage. When 
bacteria are exposed to the protein coat, or 
phage ghost, of 4>T2, all protein synthesis 
(and possibly that of RNA and DNA, too) 
is halted. The same effect is produced by 
colicin. Bacteria, however, may recover 
after exposure to phage ghosts or to colicin 
( if the latter is removed by enzymatic diges- 



tion). Sometimes, virulent phage can kill 
its host without reproducing; in such cases. 
the constituent responsible lor lethality is a 
tail-tip protein. When </>T6 is the killer, the 
lethal protein has the same X-ray inactiva- 
tion curve, specificity, and receptor site as 
colicin K. 

Since the lethal protein of T6 is very simi- 
lar to colicin K. it seems reasonable that T6 
and col K are homologous with respect to at 
least one gene. Col K can be thought of as 
a virulent phage missing that portion of the 
genome required to lyse the cell and to give 
rise to particles whose infectivity is inde- 
pendent of conjugation, yet with enough of 
the phage genome persisting to make the cell 
colicinogenic. Labeling experiments show 
that three col factors studied contained DNA 
in the amount of 4 to 7 times 10' nucleotide 
pairs — about one-tenth the amount in F or 
phage. 

During conjugation, E. coli F + cells trans- 
fer col El with high frequency, so that col 
El can exist autonomously. Autonomous 
col factors can arrive in the F~ cell as early 
as 2U minutes after conjugation is initiated. 
Since Hfr E. coli carrying col V, col I, or col 
E2 do not transfer them in linkage, these col 
factors give no evidence for an attached state. 

In Salmonella, col I in the absence of F is 
transferable via conjugation. Although cells 
carrying only col El cannot transfer it during 
conjugation, col El cells infected by col I 
transfer both col factors. Consequently, col 
I promotes the transfer of col El. When 
Salmonella harbors only col I, transfer of the 
chromosome in conjugation occurs but is 
rare. When such cells are also infected with 
col El, chromosome transfer increases one 
hundredfold. As a result, in Salmonella col 
I promotes the transfer of col El, and col El 
promotes the transfer of the chromosome. 
Col I is, therefore, a sex factor. Although 
they do not cure colicinogeny, acridine dyes 
inhibit the transfer of col factors. 

When F col~ is crossed with F~ col I, 



Bacterial Episomes and Genetic Recombination 361 

F is transferred to the F^ col I conjugant, phages, ,; F factors, RTF factors, and col fac- 
and col I is transmitted to the F + col~ con- tors. It is still too early, however, to make 
jugant with high efficiency. However, when any kind of definite suggestion as to the pre- 
F+ col I is crossed with F~ col~ , col I is cise nature of the evolutionary interrelation- 
transferred at a low rate, so that F interferes ships, if any, among these various types of 
with transfer of col I from the same parent. episomal elements. 

The preceding shows a number of simi- ,. „ , „ „ „ _ 

, . . r . , ° See also E. Seaman, E. Tarmy and J. Marmur 

lanties among temperate and virulent (1964). 

SUMMARY AND CONCLUSIONS 

A two-directional flow of F and chromosomal genetic material occurs between extra- 
chromosomal F and the chromosome. Chromosomal markers carried by free F retain 
their capacity for replication and phenotypic action. 

F', other F-merogenotes, RTF factors, and some col factors behave like episomes. 

Col factors have several characteristics in common with F and the provirus stage 
of temperate phages. Their products, colicins, resemble the lethal, tail-tip protein of 
virulent phages. Accordingly, it is possible that F, temperate phage, virulent phage, 
and col factors are related in evolution. 

Certain bacterial episomes promote their own transfer and/or transfer of other genetic 
material via conjugation. In Salmonella, col I promotes the transfer of col El which. 
in turn, promotes the transfer of the chromosome. RTF promotes the transfer of 
linked drug resistance markers and can inhibit certain promoter functions of F. 

Since they can initiate conjugation, F and its derivatives, col I, and RTF are sex 
factors. 

REFERENCES 

Adelberg, E. A., and Burns, S. N., "Genetic Variation in the Sex Factor of Escherichia 
Coli," J. Bact., 79:321-330, 1960. Reprinted in Papers on Bacterial Genetics, 
Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, pp. 353-362. 

Campbell, A. M., "Episomes," Advances in Genetics, 11:101-145, 1962. 

Clark, A. J., and Adelberg, E. A., "Bacterial Conjugation," Ann. Rev. Microbiol., 
16:289-319, 1962. 

Fredericq, P., "On the Nature of Colicinogenic Factors: A Review," J. Theoret. Biol., 
4:159-165, 1963. 

Goebel, W. F., "The Chromatographic Fractionation of Colicine K," Proc. Nat. Acad. 
Sci., U.S., 48:214-219, 1962. 

Hayes, W., The Genetics of Bacteria and their Viruses, New York: J. Wiley & Sons, 
1964. 

lacob, F., and Wollman, E. L., "Episomes, Added Genetic Elements" (in French). 
C. R. Acad. Sci. (Paris), 247:154-156, 1958. Translated and reprinted in Pa- 
pers on Bacterial Genetics, Adelberg, E. A. (Ed.), Boston: Little, Brown, 1960, 
pp. 398-400. 

Maas, R., "Exclusion of an F Lac Episome by an Hfr Gene," Proc. Nat. Acad. Sci., 
U.S., 50:1051-1059, 1963. 

Seaman, E., Tarmy, E., and Marmur, J., "Inducible Phages of Bacillus subtilis," Bio- 
chemistry, 3:607-613, 1964. 

Watanabe, T., "Infectious Heredity of Multiple Drug Resistance in Bacteria," Bact. 
Rev., 27:87-115, 1963. 



( HAPTER 27 

QUESTIONS FOR DISCUSSION 

2". I. Do .ill matings transfer F particles oi one genotype or another? Explain. 

27.2. Discuss the relationship between the transmission ol tree F particles and a seg- 
ment of the male chromosome. 

27.3. Discuss the realit) of a bacterial "chromosome" and its linear arrangement. 

27.4. B\ what series of events can you explain the origin of strain P4.\-l from P4x? 

27.5. From which particular Htr strain of E. coli could you obtain an F-Pro (pro- 
line) merogenote? How'.' 

27.6. How do \ou suppose episomes originate? 

27.7. Are integrated episomes and episomal derivatives generally able to break chro- 
mosomes? Explain. 

27.8. It has been found that c/>P2. a temperate phage which normally integrates at 
a particular chromosomal locus, position I, loses the extreme preference for 
position I when liberated from a strain carrying it in position II. How can you 
explain this finding? 

27.9. Discuss the statement: "The ability to pass from the integrated to the free state 
or vice versa is possessed by every gene in a bacterial cell." 

27. 1U. A transducing phage can carr\ two closely-linked prophages obtained from a 
doubly-lysogenic host. What conclusions can you draw with regard to the nu- 
cleotide content of a mature phage and a prophage? 

27.11. The Lac gene can either be chromosomal when integrated into the chromosome, 
or extrachromosomal when attached to free F. Should such a gene be consid- 
ered an episome? Why? 

27.12. How would you locate the position of the UV-inducible prophage of <£434 in the 
E. coli linkage map? 

27.13. How would you locate the prophage site of a noninducible (by ultraviolet light 
or zygote formation ) phage? 

27.14. Lactic dehydrogenase contains a single-strand sequence of about 33 deoxyribo- 
tides. Should this portion of the enzyme be considered genetic material? Ex- 
plain. 

27.15. During conjugation. F is reported to enhance the transfer to the F~ cell such 
substances as lactose, ultraviolet-irradiation products that induce A prophage, 
and a A repressor. Is F acting as a promoter in these cases? Explain. 



Chapter 28 

RNA AS GENETIC MATERIAL 



I 



n previous chapters, the DNA- 
containing phages were the only 
viruses discussed in detail. In 
this chapter, we study another group — vi- 
ruses that contain no DNA and are entirely, 
or mainly, ribonucleoprotein in content. 
Members of this group include many of the 
smaller viruses that attack animals (causing 
poliomyelitis, influenza, and encephalitis, for 
example), many viruses that attack plants 
(such as the tobacco mosaic and the turnip 
yellow mosaic viruses), and the small RNA- 
containing bacteriophages 1 (f2, MS2, R17. 
and others ) . These phages are all extremely 
similar, but not identical. They are the 
same size, shape, and molecular weight; they 
cross react serologically, having similar coat 
proteins; all attack only male (Hfr or F+) 
E. coli. 

The usual host for the influenza virus is 
the mammalian cell. This virus consists of a 
helical ribonucleoprotein core surrounded by 
a lipoprotein membrane. It was shown that 
the lipids in the envelope of the influenza 
virus are derived mainly from pre-existing 
lipids of the host cell, and that the composi- 
tion of the lipids varies with the strain of the 
host cell. The outer membrane of the virus 
is apparently derived from the cell membrane 
and applied when the virus leaves the cell. 
After infection by the virus, normal cellular 
growth continues for several hours. There- 

1 See T. Loeb and N. D. Zinder (1961). J. E. 
Davis and R. L. Sinsheimer ( 1963), and S. Mitra. 
M. D. Enger, and P. Kaesberg (1963). 

363 



fore, most of the RNA, protein, and DNA 
synthesized are normal cellular products and 
bear little relation to the growth of the virus. 
Using the drug, actinomycin-D, to inhibit 
normal cellular RNA synthesis, one can 
demonstrate a specific synthesis of viral 
RNA. Moreover, with the closely related 
Newcastle disease virus, which grows in the 
cytoplasm, one can show that the new (viral) 
RNA appears in the cytoplasm and not. as in 
normal cells, in the nucleus. It is probable, 
therefore, that the internal viral RNA and 
protein, as well as other viral antigens, such 
as the hemagglutinating factors, are made 
under the direction of the virus inside the 
cell. 

Several genetically-different, haploid strains 
of influenza virus have been isolated; for 
example. SWE (with markers, a c) and 
MEL (with markers. AC). When a mix- 
ture of the two strains is used to multiply- 
infect a chick's egg membranes, the mixed 
infections give progeny particles which, when 
tested, yield pure clones not only of the 
parental genotypes but also of stable recom- 
binant types {A c or a C). Since other ex- 
planations can be excluded, the results prove 
that genetic recombination occurs also be- 
tween RNA-containing viruses. 2 Genetic re- 
combination has also been demonstrated for 
the poliomyelitis virus. Whereas the occur- 
rence of genetic recombination in influenza 
may require incorporation of two or more 
pieces of viral RNA into a single particle, 
recombinant polio-virus RNA seems to occur 
in one piece. 3 Consequently, although the 
details of recombination between RNA vi- 
ruses are unknown, more than one mecha- 
nism may be involved. 

No evidence has been obtained for the 
occurrence of genetic recombination among 
viruses that attack plants. In the case of the 
tobacco mosaic virus (TMV), infection is 

- Based upon the work of F. M. Burnet and 

others. 

•See G. K. Hirst (1962). 



364 



CHAPTER 2S 



brought about experimental!} by rubbing a 
sample of virus on the leaf surface. Even 
when a high concentration of virus is used, 
onlj a small fraction of the virus particles 
(one in 10'') find and penetrate susceptible 
cells and give rise to a detectable lesion. For 





figure 28-1. Electron micrographs of to- 
bacco mosaic virus (I Ml) showing its gen- 
eral configuration (top) and its hollow core 
{middle). The bottom photo shows a particle 
whose protein has been partially removed by 
treatment with detergent, leaving a thinner 
strand of RNA. (Courtesy of R. G. Hart.) 



this reason, it is difficult to multiply-int'ect 
a tobacco cell, and experiments testing for 
genetic recombination are probably negative 
due to the lack of mixed infections. 

The TMV particle is a cylinder 3000A 
long and about 80A in radius (Figure 28-1, 
top). It has a molecular weight of about 
40 times 10 6 , of which 38 times 10 6 is pro- 
tein and 2 times 10 ,; is RNA. The outer 
dimensions of TMV are attributed to the 
helical aggregation of about 2200 identical 
protein subunits, each of which has a molec- 
ular weight of about 18,000 and contains 
158 amino acids in a single polypeptide 
chain (Figure 28-2). A cross section of 
the TMV particle shows a hollow core about 
20A in radius (Figure 28-1, middle); the 
protein subunit, therefore, adds about 60A 
to the radius. The RNA in a particle (Fig- 
ure 28-1, bottom) is typically a single, un- 
branched strand consisting of some 6400 
nucleotides threaded through the protein 
subunits at a radius of 40A. Internally the 
RNA is normally covered by about 20A and 
externally by about 40A of protein subunit. 
Since the protein subunits are arranged in a 
gently-pitched helix (49 subunits per three 
turns), the RNA forms a helix of the same 
pitch. 

When TMV in water is treated with 
phenol, the protein of the virus is extracted 
into the phenol, leaving the single RNA 
molecule intact in the water. If the tobacco 
plant is exposed to RNA molecules with pro- 
tein thus removed, the frequency of infection 
is about 500 times less than the frequency 
obtained with an equal number of whole 
virus particles; typical TMV progeny (com- 
plete with TMV protein coats) are produced. 
Repeated phenol treatments do not further 
decrease the infectivencss of the RNA. and 
no amount of protein can be detected in the 
preparations. RNasc, on the other hand, 
completely destroys the infectivencss of the 
RNA fraction but not the infectiveness of 
the whole virus. It must be concluded, 



RNA as Genetic Material 



365 



1 
Acetyl N-Ser+Tyr 


— ► Ser -► lieu— • 


5 

>Thr- 


NH 2 10 1 5 

♦Thr — ►Pro— ► Ser -►Glu -►Phe -►Vol -►Phe -►Leu — ►Ser-^Ser ->^ 


30 NH 2 

s- Ale* — Asp*— Thr 


•*-CySI- 


l-*-Asp-*- 


25 

Leu* 


NH 2 20 

— lieu-* — Leu-* — Glu-* — //«c* — Pro*-/isp*-a/o 


*-Try* 


-Ala *T 


» Leu— ►Gly — ••Asp 


NH Z 

— ►Glu- 


35 

-►Phe— » 


NH 2 

>Glu— 


NH 2 NH 2 40 ^_ 

-►Thr — ►Glu — ►Glu — ►Ala — ► Arg^-^Thr 


— ►Val- 


NH 2 

-►Glu- 


45 

♦Vol -v 


60 
s- Vol-*— Thr-*— Vol" 


NH 2 

• — Glu" 


•— Pro-* — 


55 
Ser-* 


50 nh z 
-Pro-* — Lys* — Try-*— Vol-* — Glu-*— Ser 


*— Phe- 


NH 2 

*— Glu*- 


-Arg J 


V Arg ^ Phe -►Pro 


— ►Asp 


65 

-►Ser— 1 


•Asp- 


70 ^— N 

► Phe— ► Lys ! — ►Vol — ►Tyr — »<Arg)— ►Tyr 


NH 2 

—►Asp 


— ►Ala — 


75 
♦ Vol ~V 


s- Arg*— Thr-*— Asp 


*— Phe 


*-Alo-*— 


85 

-Gly* 


80 

-Leu-* — Leu*— Ala* — Thr* — Vol-* — Leu 


* — Pro- 


• — Asp* 


-Leu*T 


»■ Asp -»(Arg)-» lieu 


— Hleu- 


95 NH 2 

-►Glu 1 


►Val- 


NH 2 NH 2 100 NH 2 

► Glu — ►Asp— ►Glu — *Ala — ► Asp -►Pro 


— ►Thr 


— »Thr- 


105 

♦Ala -v 


120 

s- Ala-*— Vol* — Thr 


•* — Ala- 


•— Asp*- 


115 

-Asp* 


t-Val-* — Arg-*— (Arg)* — Thr-*— Ala-* — Asp 


'•*— Leu 


*— Thr* 


-Glu *T 


\^ lieu— *(Arg>— ► Ser 


— ► Ala- 


125 

-►Asp— • 


■lleu- 


NH 2 130 
♦>asD-^'ei> — Weu-»VD' — ►G/tv— ►Leu 


— ►lleu- 


— *(ArgV-»Gly -v 


150 

** Leu*— Gly* — Ser' 


*— Ser- 


•-Ser*- 


145 ^-~. 140 NH 2 

-Glu-*— Phe-* — Ser* — Ser*— (Arg)*— Asp*— Tyr 


•*— Ser- 


•—Gly* 


-Thr 4f 


V Val— ►Try— ►Thr 


—►Ser 


155 

-►Gly— ► 


Pro- 


158 

►Alo — ►Thr 









figure 28-2. Amino acid sequence in the protein building block of tobacco 
mosaic virus (TMV). There are 158 amino acids in the sub-unit, the encircled 
residues indicate the points of digestion by trypsin. (Courtesy of A. Tsugita, 
D. T. Gish, J. Young, H. Fraenkel-Conrat, C. A. Knight, and W . M. Stanley, 
Proc. Nat. Acad. ScL, U.S., 46:1465, 1960.) 



therefore, that pure virus RNA is infective 
and carries all the genetic information neces- 
sary for its replication.* These experiments 
also prove that TMV protein plays no part — 
other than protecting the RNA and increas- 
ing the infectivity — in the replication either 
of the RNA genetic material or of itself. 
This conclusion is tested by what is called 
reconstitution experiments in which, under 
appropriate conditions, we can first separate 
the protein and RNA of TMV and then have 
them recombine and demonstrate the high 
infectiveness of the original virus. Using 
two genetically-different strains of this virus 
— the standard (TMV) and Holmes rib 
grass (HR) — a highly-infective virus con- 

4 RNA isolated from a number of small animal 
viruses and from coliphages f2, MS2, etc., is also 
infective. (See Vol. 27, Cold Spring Harb. Sym- 
pos. Quant. Biol., 1962, and D. E. Engelhardt and 
N. D. Zinder, 1964.) 



taining the RNA of TMV and the protein 
coat of HR can be constructed. The prog- 
eny obtained are typically TMV with both 
TMV RNA and protein coat. The recipro- 
cal construct, a virus with HR RNA and 
TMV protein, produces HR progeny typical 
in both RNA and protein. Thus, only the 
RNA of a TMV particle specifies the RNA 
and protein of the progeny virus. 5 Muta- 
tions can be induced in TMV by many of the 
agents known to be mutagenic for DNA. 
Such results and others prove that the bio- 
logical activity of the RNA depends upon its 
primary (nucleotide content) and not its sec- 
ondary (coiling pattern) structure. 

The complete amino acid sequence in the 
protein building block of tobacco mosaic 

5 The genetic experiments described for TMV are 
based largely upon work of H. Fraenkel-Conrat 
and R. C. Williams (1955), A. Gierer (1960). 
G. Schramm, and others. 



366 



( II \l- I IR 2<S 



\irus is know ii (Figure 28—2); much less is 
known about the sequence of ribotides in 
TMV. One can v lsualize TMV as a polymer 
of 3' mononucleotides ending in a nucleoside 
— shown to be adenosine.' 

Snake venom phosphodiesterase (see p. 
2S2 ) splits off 5'-nucleotides one at a time 
starting at the free 3'-OH end — called the 
.V or nucleoside end — of TMV, revealing 
that the terminal base sequence is — UACUA 
(or possibly — UAUCA). The sedimentary 
behavior of TMV RNA from which one to 
three nucleotides have been enzymatically re- 
moved does not change, implying that the 
enzyme primarily acts at the end of the RNA 
molecule as an exonuclease. The intrinsic 
integrity of the molecule is further demon- 
strated by the infectivity of such terminally- 
deleted TMV and the production of progeny 
TMV. When the progeny TMV are in turn 
treated with the same diesterase, they are 
found to release first A, then U, and then C. 
Thus, when RNA, from which several ter- 
minal nucleotides have presumably been re- 
moved, replicates, the original nucleotide se- 
quence seems to be restored in the progeny. 7 
It is possible that the sequence occurs in the 
progeny by chance; it is also possible that the 
free 5'-OH end — called the 5' or nucleotide 
end — can normally base-pair for some dis- 
tance with the 3' end of TMV. If diesterase 
does remove the terminal — CUA, the short- 
ened 3' end could be repaired by making the 
complement of a UAG — sequence (which 

8 From work of T. Sugiyama and H. Fraenkel- 

Conrat (1961). 

7 See B. Singer and H. Fraenkel-Conrat (1963). 



we would presume starts the 5' end ). On the 
other hand, it seems more likely that the in- 
fectivity of these degraded RNAs, which is 
about 109? o\' normal, is due to residual 
undegraded RNAs, which, of course, infect 
perfectly normally. The degradation with 
snake venom diesterase is very likely not 
synchronous, so that some molecules can 
have several nucleotides removed before 
others lose any. 

Results obtained with TMV and other 
RNA-viruses show that complementary 
RNA strand formation occurs during repli- 
cation of RNA genetic material. RNA- 
dependent RNA polymerase, RNA synthe- 
tase, or RNA replicase, an enzyme which 
utilizes riboside triphosphates and takes di- 
rections from RNA to make complementary 
RNA, has been isolated and purified. s One 
RNA synthetase uses the single-stranded 
RNA of the <f>MS2 — called the "plus" strand 
— as a template to synthesize the comple- 
mentary "minus 1 ' RNA strand in vivo. The 
double-stranded product — called the replica- 
tive form — is used in vitro as a natural tem- 
plate by the same or another RNA synthetase 
to synthesize "plus" strands." In this con- 
nection it should be noted that the infective 
forms of a wound virus obtained from sweet 
clover and a reovirus associated with the 
respiratory and enteric tracts of animals, in- 
cluding man, have double-stranded RNA as 
their genetic material. 10 

s By I. Haruna, K. Nozu, Y. Ohtaka, and S. Spie- 
gelman (1963), and by C. Weissmann, L. Simon, 
and S. Ochoa (1963). See D. Baltimore (1964). 
°See C. Weissmann et al. ( 1964). 
10 See P. J. Gomatos and I. Tamm (1963). 



SUMMARY AND CONCLUSIONS 

RNA is the sole carrier of genetic properties in certain viruses. Some animal RNA 
viruses can undergo genetic recombination. Mature RNA viruses carry either single- 
stranded or double-stranded RNA. Viral RNA replication involves RNA synthetase 
and the formation of complementary RNA chains. 



RNA as Genetic Material 367 

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1963. ' 
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368 < mm- i ik 28 

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QUESTIONS FOR DISCUSSION 

28.1. What conclusions can you draw from the observation that within a day or so 
alter infection with a single particle of TMV, the cell can produce about 50,000 
viral nucleic acid molecules and about 100 > 10 6 protein su